Permutations and Combinations – Part 3- Special Cases of Permutations
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 # ^{n}P_{r}= n!/ (n-r)! # Permutations of n different things taken r at a time = ^{n}P_{r} = n! / (n-r) # Permutations of n different things taken all at a time = ^{n}P_{n} = 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)}P_{r} |
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 = 4×5^{4} =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 n^{r}. |
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!
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Mujhe permutations aur combination me problem solve krne me dikkt aati hai kya aap meri help kr skti hain?
Plz