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 –
- Principle of Counting
- Permutations Introduced
- Permutations – Special cases
- Combinations – Introduction
- 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 ^{n}P_{r}.
^{n}P_{r}= 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 ^{n}P_{r} = 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 ^{4}P_{2} = 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 = ^{n}P_{r} = 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 = ^{4}P_{4} = 4! / (4-4)! = 4! / 0! = 4! = 4x3x2x1 = 24.
Permutations of n different things taken all at a time = ^{n}P_{n} = 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 ^{3}P_{2} = 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 ^{3}P_{2} = 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)}P_{r}. |
# Quickermaths Tip:
We know that ^{n}P_{r} = n! / (n-r)!Let us say we have to find out ^{12}P_{3}.
^{12}P_{3}= 12! / (12-3)! = 12! / 9! = 12x11x10x…..x1 ⁄ 9x8x7x…1 = 12x11x10 =1320
Instead of writing out so much, the moment you see ^{12}P_{3} 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 ^{12}P_{3} =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.
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