Permutation where Repetitions are not Allowed
Let’s again go to the previous example of cards (card A, B, C) to see how permutations need to be calculated when repetitions are not allowed.
- The first card could be selected 3 possible ways.
- For each selection of the first card second card could be selected 2 possible ways (since repetitions are not allowed).
- Similarly, for each selections of first and second card third card could be selected 1 possible way.
- So the total number of permutations in the case of no repetitions is:
3 X 2 X 1 = 6.
If we try to generalize the above discussion for n number of elements, the number of possible permutations for all the n items will be:
n x (n-1) x (n-2)……..3x2x1= n!
Now, let’s take the case of selecting only two cards out of three from the already discussed playing card example. Here it goes.
The first card could be selected again by 3 possible ways.
For each selections of the first card second card could be selected by 2 ways. So, total numbers of possible permutation for selecting two cards out of three will be:
If we generalize this discussion in terms of n and r (where, n is the total numbers of items and r is the number of items to be selected for each set), we can derive the formula of permutation as:
n P r = n x (n-1) x (n-2) x (n-3) x (n-r+1)
= n! / (n-r)!
I hope this has made the concept more clear to you. Practice the formulas until you have it down.