Permutations and Combinations – Part 5- Grouping & Distribution

Grouping and Distribution

This is a very important concept of permutation and combination where some higher order fundamentals of permutation and combination is involved – the reason for reserving this topic for the end.

Here is the list of articles in this series of permutation-combination for quick reference. Ideally you should check these articles in order to gain better understanding of the whole concept –

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

What is the difference between grouping and distribution?

To distribute something, first grouping is done. Only after  you have made groups of some objects, you might want to distribute these groups in various places. For example, after you made groups of some toys, you might want to distribute these groups among some children. Or, after dividing some number of toffees into groups, you might want to distribute these groups into boxes.

Just as the objects that we group can be similar or dissimilar, so can the places that we assign these groups to be similar or dissimilar.

While distributing groups, we need to keep one rule in mind: We permute the groups only if these places for distribution are dissimilar, otherwise not.

Say, we have 2 items- X and Y and I have to split them into two groups. There is only one way of doing it – X goes in one group and Y in the other. However, if I have to distribute among 2 people A and B, then these 2 groups can be permuted in 2! ways.

Division of dissimilar items into groups of EQUAL SIZE

Let’s take a very simple example. In how many ways can you divide 4 different things (say A, B, C and D) into two groups having two things each?

You would like to say that we select two things out of the four and two would be left behind, i.e. 4C2 = 6 ways. But are there really 6 ways?

Take a look. We can divide four things, A, B, C and D into two groups of two in the following ways:

AB – CD

AC – BD

AD – BC

You can keep trying but there is no fourth way to do it. So where have the remaining 3 ways calculated through 4C2 = 6 disappear? If you look carefully, there was an overlap. When we select 2 things out of 4, we can do it in 6 ways – AB, AC, AD, CD, BD and BC but when we select the first three groups, the last three get automatically selected without having to select them separately and vice-versa. So when we select AB, CD is automatically selected and vice-versa.

This overlap will manifold itself if we increase the number of items further.

So be very careful not to apply the usual combinations formula whenever we have to divide into groups of equal size. However, if the groups contain unequal number of things, then our earlier method of using combinations formula for selection will be valid as will be discussed in the next section.

 

Let me now increase the objects to 5. How would you divide these 5 distinct (dissimilar) objects into groups of 2, 2 and 1?

The single object can be chosen in 5C1 = 5 ways. The rest of the 4 objects can be divided into two equal groups in 3 ways as explained above. Therefore, total number of ways = 5 x 3 = 15.

  • The number of ways in which mn different things can be DIVIDED equally into m groups, each group containing n things = (mn)!/(n!)^m x 1/m!

  • The number of ways in which mn different things can be DISTRIBUTED equally into m groups, each group containing n things = (mn)! / (n!)^m

Note: In the distribution, order is important hence the divisible things can be arranged in m! ways since things are divided into m groups.

Division of dissimilar items into groups of UNEQUAL SIZE

Say we have k things and we have to divide them into 2 groups containing m and n things respectively such that m+n =k, then this can be done in k!/m!.n! ways. This is because m things can be selected out of k things in kCm ways and when m things are taken, n things are left to form the other group of n things which can only be done in nCn =1 way. Hence the required number of ways is kCm = m+nCm = (m+n)!/(m+n-n)!.n! = k!/m!.n!.

We can extend the same concept for increased number of groups as long as the number of items in all the groups add up to the total i.e. n.

The number of ways in which n distinct things can be DIVIDED into R unequal groups containing a1, a2, a3, ……, aR things (different number of things in each group and the groups are not distinct)

= nCa1 × (n-a1)Ca2 × … × (n-a1-a2-….-a(r-1))CaR

=n! / a1! a2! a3!…….aR!   (here a1 + a2 + a3 + … + aR = n)

 

What if there are n distinct things and we have to find out the number of ways in which they can be distributed among r persons such that some person get a1 things, another person get a2 things, . . . . and similarly someone gets aR things (each person gets different number of things)?

Number of ways in which n distinct things can be divided into R unequal groups containing a1, a2, a3, ……, aR things (different number of objects in each group and the groups are distinct)

=n! / a1! a2! a3!…….aR!x R!  (here a1 + a2 + a3 + … + aR = n)

Division of IDENTICAL / SIMILAR ITEMS into Groups

Number of ways in which n identical things can be divided into r groups, if blank groups are allowed i.e. each can receive zero or more things (here groups are numbered, i.e., distinct), where 0≤r≤n = (n+r-1)C(r-1)

 

Number of ways in which n identical things can be divided into r groups, if blank groups are not allowed i.e. each receives at least one item (here groups are numbered, i.e., distinct), where 1≤r≤n = (n-1)C(r-1)

 

Number of ways in which n identical things can be divided into r groups so that no group contains less than m items and more than k (where m<k) is coefficient of xn  in the expansion of (xm + xm+1 +…..xk)r

 

Some Important Results:

  • Number of squares in a square having ‘n’ columns and ‘n’ rows = 12 + 22 + 32 +….+ ∑n2 = n2 = n(n+1)(2n+1)/6

  • Number of rectangles in a square having ‘n’ columns and ‘n’ rows= 13 + 23 + 33 +….+ n3 = ∑n3

  • =[n(n+1)/2]2

  • Number of quadrilaterals if m parallel lines intersect n parallel lines = mC2 x nC2

 

With this discussion on Grouping and Distribution we come to an end on the topic Permutations and Combinations.

Please post in your doubts, queries and interesting problems in the comments box.

Hope you enjoyed reading these 5 articles as much as I enjoyed writing them.

Please follow and like us:

Permutations and Combinations – Part 4- Combinations

This article is the fourth in a series of 5 articles. Here we will be discussing a pet topic of all competitive examinations – Combinations. The list of 5 articles of the permutation combination series is –

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

I would advise you to go ahead with this article only after you are clear on first 3 articles in this list above.

You will be better equipped to move on with Combinations after you are thorough with the above topics. We will discuss various cases of Grouping & Distribution in the article following this.

All right, let us get started!

Difference between Permutations and Combinations

Suppose there are 3 bags (A,B and C) in my home and I want to select any 2 out of them to take with me on my holiday. In how many ways can I make the selection?

Clearly I select either AB or BC or AC i.e. 3 ways.

An important point to note is that we are talking about selection and not order here. Obviously whether I select AB or BA makes no difference.

Let us take one more example.

Suppose from a class of 10 students I have to select 3 students for a play, it is a case of Combinations. But, if I have to arrange 3 students in a line from a class of 10 students, it is a case of Permutations.

I hope I have made it clear that in permutations (rearrangement) order matters but in combinations (selections) order does not matter.

We are now in a position to define Combinations.

 Definition of Combinations

Combinations is the selection of some or all of a total of n number of things.

If out of n things we have to select r things (1≤r≤n), then the number of combinations is denoted by  nCr =  n!/r!(n-r)!

Combinations does not deal with the arrangements of the selected things. This explains division by r! which denotes the arrangement of the selected r things.

Important relation between Permutations and Combinations

‘r’ selected things can be arranged in r! ways.

So, r! x  nCr  = nPr

or,  nCr = nPr / r!

or     nCr = n! / r! (n-r)!

Example: In a class there are 6 boys and 5 girls. In how many ways can a committee of 2 boys and 2 girls be formed?

Solution: 2 boys can be selected out of 6 in 6C2 ways.

2 girls can be selected out of 5 in 5C2 ways.

So the selection can be made in 6C2 x 5C2 ways. (Product Rule: ‘and’ stands for multiplication)

Example: In a class there are 6 boys and 5 girls. A committee of 4 is to be selected such that it contains at least 1 boy and 1 girl.

Solution: There are 3 different possibilities now-

i. 1 boy and 3 girls

ii. 2 boys and 2 girls

iii. 3 boys and 1 girl

In the 1st possibility, total number of combinations =  6C1 x 5C3

In the 2nd possibility, total number of combinations =  6C2 x 5C2

In the 3rd possibility, total number of combinations =  6C3 x 5C1

But only one of the above possibilities will occur; 1st OR 2nd OR 3rd.

So the total number of required combinations is  6C1 x 5C3 +  6C2 x 5C2 + 6C3 x 5C1

Some Important Results on Combinations

  • nCr = nCn-r                                           (0≤r ≤n)
  • nC0 = nCn = 1
  • nCr + nCr-1 = n+1Cr                  
  • If nCp = nCq , then p = q or p + q = n (p,q € W)

Restricted Combination

The number of combinations of ‘n’ different things taken ‘r’ at a time  subject to restriction that p particular things –

i) will never occur = n-pCr

ii) will always occur = n-pCr-p

Number of ways of selecting one or more things from a group of n distinct things = nC1 + nC2 + nC3 + …… +  nCn  = 2n – 1 .

Number of ways of selecting zero or more things from a group of n distinct things = nC0 + nC1 + nC2 + nC3 + …… +  nCn  = 2n

With this we conclude the topic Combinations. Feel free to post interesting questions which we all can share and solve together. In the following article we will discuss the rules on Grouping & Distribution- a very important concept of permutation and combination.

Please follow and like us:
financial calculator

Which Financial Calculator to Buy?

Are you taking any financial certification exams like CFA, CFP or CPA?

Are you having a tough time calculating your loan payments?

Are you facing difficulties figuring out interest amounts and payment schedule?

Are you interested in calculating cash expenses and making other financial projections for your business?

If the answer to any one of the above mentioned questions is a yes, you might consider buying a financial calculator for yourself.

A financial calculator is a highly powerful calculating machine with numerous additional features and calculation capabilities as compared to the simple add-subtract-multiply-divide functions of a standard calculator. It essentially works like a mini computer. Some of the higher-end models of financial calculators also include graphing capabilities that enables you to calculate future growth rates based on current mathematical estimates or to store formulas inside the built-in data drives.

Based on functionality, price considerations and exam acceptance, I am presenting here a list of the best financial calculators to choose from.

HP 10BII

The HP10BII is one of the most affordable financial calculators available and thankfully, you do not need to compromise on quality on this one. It is easy to use and learn and a true value for money. Of course there are more powerful financial calculators out there but not one is easier to use with all the features that you’ll need as a beginner. It is capable of completing over 100 different math functions and it has special programs for bond valuations and bond yield. Unless you’re a financial expert with specific needs, you can definitely look at this powerful equipment from HP.

The HP 10BII+ takes everything that is great about the HP 10BII and then simply extends on the functionality. Buy Online

CFP Standards: The Certified Financial Planning Board (CFP) has strict regulations regarding acceptable calculator models for students taking the exam. It does not accept any calculator with alphabetical keys. HP 10BII is one of the few calculators allowed for the exam. Therefore, I would strongly recommend this particular model to students who are planning to take this certification test in the future.

Texas Instruments BA II Plus Financial

There are two versions of this particular calculator. The Texas Instruments BAII Plus Financial version is the scaled-down model of the professional one. It is best for students who have just started with preparations for financial certification examinations. The better versions can always be bought later as you progress in your career path. Its price is probably one of the best reasons to select this calculator, especially for those on strict budgets.

The calculator features finance functions such as amortization, bonds, cash flows (up to 24 uneven), depreciation, and more.

CFA & CFP Standards: Only two calculator models are authorized by the CFA Program exams and one of these is the BAII Plus. It is also acceptable for the CFP certification exam.  Buy Online

HP 12C

HP 12 C is the ideal purchase for students and professionals alike. Price wise it strikes just the right balance. It is more expensive than the budget financial calculators and cheaper than the high-end graphic calculators.

The large layout of buttons and 10-digit display screen is perhaps its biggest draw making it extremely easy to use. In terms of functionality, it has over 120 business and finance functions such as, cash flow, depreciation, bonds, loan payments, amortization, and more. Overall, a great calculator except its price. There are less expensive financial calculators that perform the same functions at a lower price but for banking, real estate, and insurance professionals it may be worth the extra money.

CFA Standards: This is the other calculator acceptable for use in CFA certification apart from BAII Plus discussed above. It is also acceptable for use in CFP certification examinations. Buy Online

Texas Instruments BAII Plus Professional Financial

Slightly more expensive than the entry level Plus II, the BAII Plus Professional is surely a power performer. In addition to the basic features, it comes with many advanced functions for trignometric and logarithmic (including natural log) calculations . It is also an ideal buy for students who need to incorporate advanced calculus.

Though it is pricier than the standard BAII Plus, it definitely is an intelligent buy for students looking for an accounting or economic career as it will last you long.

CFA & CFP Standards: It is acceptable for use in both CFA and CFP certification examinations. Buy Online

HP 17 BII

The HP 17BII can be easily compared to the HP 12C both in terms of performance and price. The main difference is the display screen. While 12C has a single line 10-digit large display, 17BII has a split screen display that allows users to see two different numerical lines. That is not all. The 17BII has additional 250 finance functions as compared to the HP 12c. One of these is the dual variable forecasting feature, which is surely a big plus point as compared to other budget and moderately-priced calculators. It is helpful for statistics classes, but obviously if your pocket permits, a graphic calculator will serve a better purpose there.

CFP Standards: This particular calculator is acceptable for the CFP exam. Buy Online

 Texas Instruments TI-84 Plus

Essentially a graphic calculator, TI 84 is an improved version of the TI 83. It incorporates graphical and statistical programs in addition to all the regular functions of a financial calculator. Fact is fact, it is a very expensive calculator. So it suits only those people who actually need the added graphical and statistical capabilities of this power machine. It takes a while to get accustomed to this one because of the many many features.

Like most graphic calculators, it has a display screen large enough to show an entire sequence of numbers as well as the display of graphs.

Certification Exam Standards: It is forbidden for use in any financial certification exam! Buy Online

This article contains some affiliate links to flipkart.com.

Which calculator do you use and recommend?

Please follow and like us:

Permutations and Combinations – Part 3- Special Cases of Permutations

There are 5 articles in this learning series of permutation-combination. Here’s the list –

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

Since you are already familiar with the fundamental principle of counting and the concept of permutations, we can move ahead with some special cases of permutations in this article.

By now you already know:

# With respect to fundamental principle of counting, ‘and’ stands for multiplication & ‘or’ stands for addition

# nPr= n!/ (n-r)!

# Permutations of n different things taken r at a time = nPr = n! / (n-r)

# Permutations of n different things taken all at a time = nPn = n!

# Permutations of n different things taken r at a time, when one particular thing always occurs = r.(n-1)P(r-1)

# Permutations of n different things taken r at a time when a particular thing never occurs = (n-1)Pr

Let us begin with a question.

 In in how many ways can the letters of the word WATER be arranged so that we have a new pattern every time?

This is permutation of n different things taken all at a time which is equal to n!

Hence, total number of different arrangements possible is 5! =120.

Another way to look at it is we have 5 places to be occupied by 5 different letters.

The 1st place can be filled by any of the 5 letters, hence 5 ways. The 2nd place can be filled by any one of the remaining 4 letters as one letter has already been fixed at the first place, hence 4 ways. Similarly, the 3rd place can be filled in 3 ways and the 2nd in 2 ways. The 5th place can be filled in only one way as there is no choice but to fill it by the remaining 1 letter.

So going by the product rule, this can be done in 5x4x3x2x1 = 120 ways.

Permutation of n things when some are identical or Permutation of n things not all different:

What happens when we have to find out the number of permutations when certain items are identical?

If 2 exactly similar red chairs(R1 & R2) and 1 black chair(B) are to be arranged, then please note that one cannot distinguish between the 2 red chairs. This is to say that there is no difference between R1 B R2 and R2 B R1 because they will both look the same as I cannot differentiate between R1 and R2 as they are exactly same.

So the total no. of arrangements possible will be 3! / 2! = 3. They will be BRR RBR RRB.

We have divided by 2! to take care of the ‘two’ items that are same.

If out of n things, p are exactly alike of one kind, q exactly alike of second kind and r exactly alike of third kind and the rest are all different, then the number of permutations of n things taken all at a time                  = n! / (p!q!r!)

 Example: In how many ways can the letters of the word COMMITTEE can be arranged

i. using all the letters

ii. if all the vowels are together

 Solution:

i. Total letters = 9 and identical letters are 2M 2T and 2E.

So total no. of arrangements = 9! / 2!2!2!

ii. Since all vowels must appear together we consider them as one unit. There are 4 vowels- O I E E. So now we have 5 letters. Out of these we have 2M and 2T. These 5 letters can be arranged in 5! / 2! 2! ways.

In the group of 4 vowels, the 4 vowels can arrange themselves in 4!/2! ways.

So total no. of words formed = 5!/2! 2! X 4!/2!

Permutations where repetitions are allowed

While dealing with letters and digits, you will often come across cases where repetition in permutation is allowed or not allowed. You have to be very careful as to what is asked for because the treatment for both the cases is absolutely different.

Example: How many numbers of 5 digits can be formed with the digits 0,1,2,3,4

i. if the digits cannot repeat themselves

ii. if the digits can repeat themselves

 Solution:

i. The 1st place (ten-thousandth place) can assume only non-zero digits. Hence it can be occupied in 4 ways. The 2nd place can be occupied by any of the remaining 4 digits, i.e. 4 ways. Similarly, the 3rd, 4th and 5th place in 3, 2 and 1 ways respectively.

Total no. of numbers formed = 4x4x3x2x1 = 96

ii. The 1st place (ten-thousandth place) can assume only non-zero digits. Hence it can be occupied in 4 ways. Since repetition is allowed, the 2nd, 3rd, 4th and 5th places can all be filled in 5 ways each i.e. we have a choice of 5 digits (0,1,2,3,4) for each place.

So total no. of numbers formed = 4x5x5x5x5 = 4×54 =2500

The number of permutations of n different things taken r at a time, when each may be repeated any number of times in each arrangement is nr.

Geometrical arrangements

Circular permutation:

Sitting in a circle is not the same as sitting in a straight line. A circle does not have any starting point or ending point. Thus in a circular permutation, one thing is kept fixed and the others are then arranged relative to this fixed item. Then it is treated like a linear arrangement.

The number of circular permutations of n different things taken all at a time = (n-1)!

Fix any one as reference point, the remaining other n-1 things can be arranged in (n-1)! ways.

What if we are taking into consideration beaded necklace or a garland wherein clockwise and anticlockwise arrangements are the same?

We simply divide (n-1)! by 2 to take into account the two same clockwise and anticlockwise arrangements.

If the clockwise and anticlockwise orders are not distinguishable, then the number of permutations = (n-1)! / 2

Arrangement around a regular polygon:

If n people are to be arranged around a p sided regular polygon, such that each side of the polygon contains the same number of people, then the number of arrangements possible is n!/p.

For example, 15 people are to be arranged around a pentagon shaped table having 3 people on each side of the table, number of arrangements will be 15!/5.

Please note if the polygon is not regular, then the number of arrangements will be n! irrespective of the sides of the polygon.

 Special case of arrangement around a rectangular table:

Rectangle is a special case because though it is not a regular polygon, it is a symmetrical quadrilateral with opposite sides equal. So, if n people are to be arranged around a rectangular table, such that there are the same number of people on each of its 4 sides, then the total number of arrangements possible is n!/2. Here 2 signifies the degree of symmetry of the rectangle.

 

With this we have completed all the concepts of permutations. In the following two articles, we will discuss combinations and rules for grouping and distribution.Feel free to ask any doubt or question that arises in your mind after reading this article.

 

Keep the comments coming. Your feedback is very important.

Till then, happy permuting!

 

Please follow and like us:

Permutations and Combinations – Part 2- Permutations Introduced

This article is the second of a series of five articles. In the first part we had discussed the fundamental principle of counting. The first part is essential for you to understand the difference between ‘and’ & ‘or’ in the realm of permutations and combinations. We will discuss the concept of permutations in this article and permutations- special cases, combinations and Grouping & Distribution in the three articles following this.

Here is the list of articles in this series of permutation-combination for quick reference –

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

Permutations

The arrangements of a given number of things taking some or all of them at a time are called permutations.
For example, the permutations of three alphabets x, y, z taken two at a time are xy, xz, yx, yz, zx, zy.
A point to be noted is that arrangement or order is very important in permutations. Hence xy is distinctly different from yx.

If r things are taken at a time out of a total of n things, then the total number of permutations is denoted by nPr.

nPr= n!/ (n-r)!

Now you will ask why is this so. Let’s clear this.
First object can be selected in ‘n’ ways. Second object can be selected in (n-1) ways. Third object can be selected in (n-2) ways. Similarly the rth object can be selected in (n-(r-1)) = (n-r+1) ways. Therefore the total number of ways of arranging these ‘r’ objects = n x (n-1) x (n-2) x (n-3) x ……x (n-r+1)
={n x (n-1) x (n-2) x ……(n-r+1) x (n-r) x (n-r-1) x…. 3 x 2 x 1} / {(n-r) x (n-r-1) x…..x 1}= n! / (n-r)!

Hence nPr  = n! / (n-r)!

Let us now understand further with the help of different examples.

Example 1: There are 4 boxes. Find the total number of arrangements if we can arrange only 2 boxes at a time.

Solution: Out of 4 boxes, we are arranging 2 at a time.
So total number of arrangements possible is 4P2 = 4! / (4-2)! = 4! / 2! = 4x3x2x1 / 2×1  = 12
Let us verify. Let us name the boxes A, B, C, D.
Total number of arrangements possible are AB, BC, CD, BA, BC, BD, CA, CB, CD, DA, DB, DC.

Permutations of n different things taken r at a time = nPr = n! / (n-r)!

Example 2: In the above example, what if all the 4 boxes are selected at a time? How many arrangements are possible then?

Solution: Total no. of arrangements possible = 4P4 = 4! / (4-4)! = 4! / 0! = 4! = 4x3x2x1 = 24.

Permutations of n different things taken all at a time = nPn = n!

Example 3: If out of the 4 boxes, one particular box should always be selected; then how many arrangements are possible if 3 boxes are selected at a time?

Solution: Since one box should always be selected we have to select 3-1 boxes out of 4-1 boxes. This can be done in 3P2 = 3! / (3-2)! = 3! / 1! = 3x2x1 / 1 = 6 arrangements.
With each of these 6 arrangements our preselected box can be arranged in 3×6 = 18 ways.

Wondering how?

Let us name these boxes A, B, C and D and D has to be always present.
So now A, B and C can be arranged as AB, AC, BA, BC, CA, CB.
With AB, D can be arranged as DAB, ADB, ABD i.e. 3 ways.
D can be arranged with the remaining 5 arrangements similarly.

Hence in total there can be 18 arrangements.

Permutations of n different things taken r at a time, when one particular thing always occurs = r.(n-1)P(r-1)

Example 4: How many arrangements are possible if out of the 4 boxes – A, B, C and D one particular box D is never selected, taken 2 at a time?

Solution: Since D is never to be selected, we have to take into account A, B and C.
We can arrange A, B and C taken 2 at a time in 3P2 = 3! / (3-2)! = 3! / 1! = 3x2x1 / 1 = 6 ways. i.e. AB, AC, BA, BC, CA, CB.

So when one particular item is never chosen, we just ignore it and treat the problem as if that particular item is not present in the total number of items.

Permutations of n different things taken r at a time when a particular thing never occurs = (n-1)Pr.

# Quickermaths Tip:
We know that nPr = n! / (n-r)!

Let us say we have to find out 12P3.

12P3= 12! / (12-3)! = 12! / 9! = 12x11x10x…..x1 ⁄ 9x8x7x…1 = 12x11x10 =1320

Instead of writing out so much, the moment you see 12P3 you should know that you have to multiply 3 numbers.

Starting from 12, we take in 3 numbers in the descending order and multiply them out.

Learn to get into the habit of writing 12P3 =12x11x10 straightaway.

This helps in faster calculation.

Hope the concept of permutations is clear to you by now. We shall discuss some special cases of permutations including circular permutations in the article following this after which we will move on to Combinations and Grouping & Distribution.
Keep your comments and other interesting problems coming.

Please follow and like us:
Squaring made easy

Squaring Numbers in a Simple and Fun way

Squaring numbers is a tedious job if you do not know vedic maths shortcuts and other tricks and tips. Imagine having to waste your precious seconds in carrying out actual multiplications to find out squares during an exam. It can prove disastrous.

It is always advisable to learn various time saving tricks and alternate methods to find out percentages, fractions, squares, cubes, etc. because these will be useful in Quantitative Aptitude as well as Logical Reasoning sections of any competitive examination you take.

The idea of this post is contributed by Piyush Goel and it’s further edited by me to share with you a trend that he observed in squares of numbers.

If we talk about Square, Square is the result of multiplying a number by itself. So, square of 3 is 3×3=9. Similarly, square of 10 is 10×10=100. Since you know what is a square, observe this:

If we have to square 11, instead of 11×11=121,
we can simply put 1 (1*2)(1^2) and get 121.
Similarly, we can write the square of 12 as: 1(2*2)(2^2) and get 144. In the same fashion we can get the square of 13 as 1(3*2)(3^2)=169 and the square of 14 as 1(4*2)(4^2)=1|8|16 =196(after carrying over the 1 of 16 and adding it to 8).

Look at (2,4,6,8,10,12,14,16,18,20….difference of 2) & (1,4,9,16,25,36,49,64,81…….. difference is 3,5,7,9,11,13,15,17,19….) and difference of 3,5,7,9,11 is 2 so there is True Symmetry of 2.

11^2 = 1 2 1
12^2 = 1 4 4
13^2 = 1 6 9
14^2 = 1 8 16 = 100 + 80 + 16 = 196
15^2 = 110 25 = 100 + 100 + 25 =225
16^2 = 112 36 = 100 + 120 + 36= 256
17^2 = 114 49 = 100 + 140 + 49= 289
18^2 = 116 64 = 100 + 160 + 64= 324
19^2 = 118 81 = 100 + 180 + 81= 361
20^2 = 111^2 = 1 20 100 = 100 + 200 + 100 = 400
21^2 = 111^2 = 1 22 121 = 100 + 220 +121 = 441
22^2 = 112^2 = 1 24 144 = 100 + 240 +240 = 484
23^2 = 113^2 = 1 26 169 = 100 + 260+169 = 529
24^2 = 114^2 = 1 28 196 = 100 + 280+196 =576
25^2 = 115^2 = 1 30 225 = 100 +300 +225 =625
26^2 = 116^2 = 132 256 = 100 + 320 +256 =676
27^2 = 117^2 = 134 289 = 100 + 340 + 289 =729
28^2 = 118^2 =136 324 = 100 + 360 + 324 = 784
29^2 = 119^2 = 138 361 = 100 + 380 + 361 = 841
30^2 = 120^2 = 1 40 400 = 100 + 400 +400 =900
31^2 = 121^2= 121^2 = 1 42(111^2) =1 42 (1 22 121) =100 + 420 + 441 =961
41^2 = 131^2 = 1 62 (1 21^2) = 1 62 (1 42) (1 11^2) = 1 62 (1 42) (1 22 121) = 1 62 (961) = 100 + 620 + 961 = 1681

This is time saving and the best part is we do not have to mug up additional formulae or large algorithms of processes.

Please share your feedback and other interesting ways that you know.

About The Author :
Piyush Goel, born on 10th February, 1967. He has a Diploma in Mechanical Engineering and a Diploma in Vastu Shastra. Always wanting to do something new, Mr. Goel has written 15 Spiritual and World Famous Books with his own hands in Mirror Image in different ways. He is now known as “Mirror Image Man”. He also has World’s First Hand-written Needle Book “Madhushala” to his credit.

Edited by Preeti Patawari

Please follow and like us:

Permutations and Combinations – Part 1- Fundamental Principle of Counting

Permutations and Combinations is one of the most logical phenomenon of mathematics wherein there are no formulae to mug up. Rather, it tests your ability to understand the problem and interpret the situation logically. It is more of application of common sense. That is why you will see that most questions can be solved without actually knowing the technicals of permutations and combinations. Great news, isn’t it? 😉

Since it is an extensive topic, I am going to break this article into 4 parts (click on the links to read further)-

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

Before proceeding further, let us quickly define Factorial.

Factorial of a number or n! is the product of n consecutive natural numbers starting from 1 to n. Factorial word is represented by ‘!’ or ‘L’. Hence, 4! is 1x2x3x4 = 24.

Note: Factorial of zero or 0!=1

Fundamental Principle of Counting

Product Rule

If one operation can be done in x ways and corresponding to each way of performing the first operation, a second operation can be performed in y ways, then the two operations together can be performed in xy ways.

If after two operations are performed in any one of the xy ways, a third operation can be done in z ways, then the three operations together can be performed in xyz ways.

Let us take an example.

permutation combination

A, B, C and D are four places and a traveller has to go from A to D via B and C.

He can go from A to B in 4 ways and corresponding to each way he can take any one of the 2 ways to reach C. Hence A to C can be reached in 4×2=8 ways.

Corresponding to each of these 8 ways of reaching C from A, there are 3 ways to reach D and the traveller can choose any one of them.

Hence, A to D can be reached in 4x2x3=24 ways!

Here the different operations are mutually inclusive. It implies that all the operations are being done in succession. In this case we use the word ‘and’ to complete all stages of operation and the meaning of ‘and’ is multiplication.

One more example before we move ahead.

A tricolour flag is to be formed having three horizontal strips of three different colours. 5 colours are available. How many differently designed flags can be prepared?

Solution: First strip can be coloured in 5 ways, second strip can be coloured in any of the remaining 4 colours, and the third strip can be coloured in any of the remaining 3 colours.

Hence, we can get 5x4x3 = 60 differently designed flags.

Addition Rule

If there are two operations such that they can be performed independently in x and y ways respectively, then either of the two jobs can be done in (x+y) ways.

Let us take the example of four places A, B, C and D taken above.

permutation combinationThere are 4 different roads from B to A and 2 different roads from B to C. In how many ways can a person go to A or C from B? The answer is 4+2=6 ways.

Here, the different operations are mutually exclusive. It implies either of the operations is chosen. in this case we use the word ‘or’ between various operations and the meaning of ‘or’ is addition.

The product rule and the addition rule signify the cases of ‘and’ & ‘or’.

Deciphering between these ‘and’ & ‘or’ is the main concern of most questions of permutations and combinations. So you have to read the questions very carefully and keep answering this question in your mind – Do I have to choose between the two jobs or perform both in succession? If you can answer this correctly which of course you will be able to with careful reading and interpretation and some practice, you have already won the battle!

Stay tuned for conceptual clarity on permutations (introduction & special cases) and combinations. Keep the comments coming, folks. 

Please follow and like us:

Probability – Concept Explained

I received quite a few requests for putting up a write up on probability. Today I finally decided to pen it down.

Probability tells how likely something is to happen. It tells the chance that an event will occur.

If I have to toss a coin, I cannot be certain that a head or a tail will occur. I can only talk about what is the chance for a head or a tail to occur.

Assuming that the coin will fall on either side only and not stand vertical ;), there are only two possible outcomes when I toss a coin – head or tail. Therefore, both are equally likely (unless of course the coin is biased). probability of getting a head is ½ and probability of getting a tail is also ½. In other words there is a 50-50 chance for both the outcomes.

When a single die (Yes, die. Actually dice is the plural form of die) is rolled, there are 6 possible outcomes – 1, 2, 3, 4, 5 & 6 and all the six outcomes stand an equal chance of occurring. So each outcome has a probability of ⅙.

In general, probability of an event = no. of favourable outcomes/total no. of possible outcomes

0≤P(E)≤1

It might have occurred to you by now that probability of any event will always lie between 0 and 1. 0 being impossible and 1 being absolutely certain.

For example, probability of getting a 9 on throwing a die is 0 because it is impossible (at least in the case of a normal die) and probability of getting a number less than 10 is 1 because it is certain.

Determining the total number of outcomes

If we have to toss 1 coin there are only two outcomes.
If we toss 2 coins together, there are 4 outcomes – HH TT HT TH
If we toss 3 coins together, then?
HHH TTT HHT HTH HTT TTH THT THH – 8 outcomes.

So how do we compute this?

Every time you increase a coin, the total number of outcomes doubles.

What if it is a die? Rolling 1 die results in 6 outcomes; 2 die in 36 outcomes; 3 die in 216 outcomes. Phew!!

Can we generalize this? Yes we can.

Total number of outcomes = (Number of outcomes per object) to the power  Number of objects
That means if i am tossing 8 coins together, total no. of outcomes that I will get is 2 to the power 8 (2^8)
Similarly, total no. of outcomes on rolling 10 die will be 6 to the power 10 (6^10).

Odds in Favour and Odds Against

You will find ample usage of these two terms in probability related questions. Odds in probability defines a ratio between favourable and unfavourable events.
Odds in favour = No. of favourable cases / No. of unfavourable cases
Odds against = No. of unfavourable cases / No. of favourable cases
Example: Two fair coins are tossed. Find the odds in favour of getting at least one head.
Solution: No. of favourable cases= 3 (HT TH HH)
No. of unfavourable cases= 1 (TT)
Hence, Odds in favour of getting at least one head ( i.e. one or more) = ¾.

Mutually exclusive and exhaustive system of events

If two events are said to be mutually exclusive then if one happens, the other cannot happen and vice-versa. Basically, the events cannot occur simultaneously.

In general P(A or B) = P(A) + P(B) – P(AB)

If two events are mutually exclusive then P(AB)=0

For example, in rolling a die

A: the event that the number is odd
B: the event that the number is even
C: the event that the number is a multiple of 3

Let us see what each of these events encompass.
A: {1 3 5} B: {2 4 6} C: {3 6}

Events A and B are mutually exclusive because a number can be either odd or even and not both simultaneously. Basically, there is no common element between A and B.

Events A and C are not mutually exclusive or disjoint since they have a common outcome – 3.

Events A{1,3,5} and B{2,4,6} together also form an exhaustive system of events as together they have included all possible outcomes and there is no outcome beyond A and B for the event ‘rolling a die’.

Additional Law of Probability

In case of mutually exclusive events, probability that either event A or event B will occur in a single trial is given by: P(A or B) = P(A) + P(B). This can also be compared with set theory.

Similarly, P(neither A nor B) = 1- P(A or B).

In case of 3 events,
P(A or B or C) = P(A) + P(B) + P(C) – P(A and B) – P(B and C) – P(A and C) + P(A and B and C)
If A, B and C are mutually exclusive then P(A and B) = P(B and C) = P(A and C) = P(A and B and C) = 0

Independent Events

Two events are independent if the occurrence of one has no effect on the occurrence of the other.

For example, on rolling a die and tossing a coin together:
A: the event that the number 5 turns up
B: the event that tail turns up
Here both A and B are completely independent of each other. Getting a 5 cannot have any impact on getting a tail.

Take another example.
Suppose a bag contains x red balls and y black balls and two balls are drawn out.

There can be two cases here –
one, the first ball is replaced and then the second ball is drawn;
two, the first ball is not replaced and the second ball is drawn out of the remaining balls.

If the ball drawn in the first draw is not replaced back in the bag, then two events of drawing the ball are dependent because first draw of the ball would determine the probability of drawing the second ball. If the ball drawn in the first draw is replaced back in the bag, then two events are independent because first draw of a ball will have no effect on the second draw.

Multiplication law of probability

If the events A and B are independent, then P(A and B) = P(A) x P(B).

Conditional Probability

The concept of conditional probability is used only in case of dependent events.

Let A and B be two dependent events, then probability of occurrence of event A when B has already occurred is given by P(A|B) = P(AB)P(B).

Let us understand this with an example. A pair of dice is thrown simultaneously. We have to find the probability that the sum obtained is 8 when there is an even number on the 1st die.

A: Sum of 8 is achieved.

A= {(2,6), (3,5), (4,4), (5,3), (6,2)}

B: we get an even number on the 1st die.

B= {(2,1), (2,2), (2,3),…,,(4,1), (4,2), (4,3),…, (6,1), (6,2), (6,3),…(6,6)}

Total no. of outcomes= 62=36

P(A)= 5/36 and P(B)=18/36

P(A|B) = P(AB)P(B)= (2,6) (4,4) (6,2)P(B)= 3/3618/36=3/18.

I have covered all the basics that you need to know to understand thoroughly the basics of probability.

This article is written from the viewpoint of competitive exam takers. Hence I have put in what they need to know. I have avoided detailed definitions of every single term to maintain crispness. Also I have deliberately avoided explanations using too much of set theory and series and functions notations to keep it simple and easy to digest.

Hope you like it. Waiting for your feedback!

Please follow and like us:

How I chanced upon Factorial Function

Today we will see how Piyush Goel(the contributor of this article) discovered something while playing with numbers. Here is the story contributed by him:

One day while sitting idle, and having nothing important to do I was just scribbling on paper. I was writing random numbers and thinking how I could link them. To my amazement I discovered this:

I wrote down 0, 1, 2, 3, 4, 5.

Next to each number I wrote their respective squares, viz. 0,1, 4, 9, 16, 25.

Then start subtracting each successive square from the next bigger square.

It looks something like this: (1-0), (4-1), (9-4), (16-9) and (25-16).

What is the result?

I got 1, 3, 5, 7, 9.

Now again subtract each successive number from the next one in the order.

We are performing this: (3-1), (5-3), (7-5) and (9-7).

Surprisingly, I got (2, 2, 2, 2). A 2 in each case. I was amazed and decided to take a step further. This time with cubes.

So here go my numbers 0, 1, 2, 3 ,4, 5, 6.

And their respective cubes: 0, 1, 8, 27, 64, 125, 216.

Time to perform first round of successive subtraction, i.e. (1-0), (8-1), (27-8), (64-27), (125-64) and (216-125). The result: 1, 7, 19, 37, 61, 91.

Second round of successive subtraction as performed earlier: (7-1), (19-7), (37-19), (61-37) and (91-61).

And this is what I got: 6, 12, 18, 24, 30.

Successive subtraction performed the third time: (12-6), (18-12), (24-18) and (30-24).

And this time I got the result (6, 6, 6, 6).

You see, squaring means ‘to the power 2’ and I got the result 2 after 2 successive subtractions, which is the value of 2!

Cubing means ‘to the power 3’ and after performing 3 successive subtraction I got 6 in each case which is nothing but the value of 3!

I similarly performed with 4th and 5th powers of the first few whole numbers. I was surprised to find the final value of 24 and 120 which are of course the values of 4! and 5!.

It was all right up to 5, but I wanted to go a notch higher and check for 6 just for my personal satisfaction. While doing for 6th power the calculations were massive and numbers so huge that it took me a couple of hours to do my calculations. Just to be sure I also did it for numbers upto 15.

Put 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15 and their respective 6th powers:

0,1,64,729, 4096, 15625, 46356, 11649, 262144, 531441, 1000000, 1771561, 2985984,4826809, 7829536 and 11390625.

I kept on doing successive subtractions till to my contentment I finally got 720 for all which is the value of 6!

With little efforts, we can work out algebraic relationship for the above but I’ll leave it for some other time. Do share your comments.

About The Author:

Piyush Goel, born on 10th February, 1967. He has a Diploma in Mechanical Engineering and a Diploma in Vastu Shastra. Always wanting to do something new, Mr. Goel has written 15 Spiritual and World Famous Books with his own hands in Mirror Image in different ways. He is now known as “Mirror Image Man”. He also has World’s First Hand-written Needle Book “Madhushala” to his credit.

Please follow and like us:

How to find the largest power of a number that divides a factorial number?

Factorial

Factorial of a number or n! is the product of n consecutive natural numbers starting from 1 to n. Hence, 4! is nothing but 1x2x3x4 = 24

Common Type of Questions on Factorials

Two types of questions are very common on factorial.

One, finding out the number of zeros at the end of a factorial expansion. You can find the detailed explanation on how to deal with such questions here.

Two, finding out the largest power of any number that divides any factorial number.
Today, I am going to explain how to approach questions like

Find the largest power of 5 contained in 124!
Find the highest power of 7 that can exactly divide 777!

Let us get started.

Let’s say we have to find out the largest power of 5 contained in 25!
1!, 2!, 3! and 4! are not divisible by 5 because 5 is not a factor in these factorial numbers.

Now if I consider any factorial number greater than 4!, every number consists of 5 or the higher powers of 5.
5!= 1x2x3x4x5
6!= 1x2x3x4x5x6
… …. … …. …
10!= 1x2x3x4x5x6x7x8x9x10
…. …. …. …. …. ….
25!= 1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20x21x22x23x24x25

Multiples of 5 in the expansion of 25! are 5, 10, 15, 20 and 25.

Basically, 25 divided by 5 equals 5. So there are 5 multiples of 5 from 1 to 25.

But, 25=5X5. This means that 25 is giving me an extra 5 which I need to account for. With this logic every multiple of 25 will give me an extra 5.

25 divided by 25 equals 1. So there is 1 multiple of 25 from 1 to 25.

Therefore highest power of 5s in 25! = 25/5+25/25= 5+1 = 6

The same reasoning extends to larger numbers. Suppose we had to find the highest power of 5 in 1000!

There are 1000/5=200 multiples of 5 between 1 and 1000.

The next power of 5 is 25 and there are 1000/25=40 multiples of 25 between 1 and 1000.

The next power of 5 is 53=125 and each such multiple of 125 gives me still one more extra 5 as compared to other multiples of 25. There are 1000 /125 = 8 such multiples of 125 between 1 and 1000.

The next power of 5 is 54=625 and there is only one multiple of 625 between 1 and 1000, 625 itself.
Hence the total number of 5s in 1000! = 200 + 40 + 8 + 1 = 249.

You keep on dividing with higher powers of the given number till you get 1 as the whole number part. And of course, while adding up we consider only the whole number part and ignore the fractional part.

If you observe, this is nothing but the greatest integer function.
We have to find out the highest power of k that can exactly divide n!.

We divide n by k, n by k2, n by k3.. and so on till we get n by kx equals to 1 or 1 and some decimal part ;(where [P] means the greatest integer less than or equal to P) and then add up.

Highest power of prime number k in n! =[n/k]+[n/k2]+[n/k3]+…….+[n/kx]; where [P] means the greatest integer less than or equal to P.

Example: Find the highest power of 2 in 50!
Solution: Highest power of 2 in 50! = [50/2]+[50/4]+[50/8]+[50/16]+[50/32]= 25+12+6+3+1= 47

Example: Find the highest power of 30 in 40!
Solution: Express 30 as product of its prime factors. 30=2x3x5. So to make a 30 you need each of 2, 3 and 5. Now in 40! there will be more 2s compared to 3s and more 3s compared to 5s. Hence to extract each 30 out of 40!, we should be bothered only about number of 5s in 40!.

Memory trick: Find number of the largest prime factor of 30 i.e. 5 in 40!.

Highest power of 5 in 40! = [40/5]+[40/25] = 8+1=9.

What if I am asked to find the highest power of numbers like 17, 19 or say 23? Then I will have to deal with numbers like 173,194, 235, etc. Finding the higher powers of such divisors and the subsequent division can be a real pain. So is there an alternate method? Luckily, yes.

Short-cut to find the highest power of a number in a factorial

By now it is clear that we consider only the integral part of the quotient while performing the series of divisions. So we can do this calculation in an easier method. We just divide the number and divide the succeeding quotients by the same number till we get 0. Finally add up all the quotients.

In this method we get rid of finding higher powers of the divisor.
So if we have to find the highest power of 5 in 1000!

200+40+8+1= 249. Isn’t this just awesome. I no longer have to worry about the values of higher powers of 5. Also division by 5 is relatively easier than division by 25, 125 or 625.

Share your feedback and let me know if you liked this.

Please follow and like us: