Case 4: 2 letters Repetition is allowed
In this case, bucket 1 can have any of the 4 letters“ C, A, R, D”. That is 4 choices
Bucket 2 can also have any of the 4 letters, since repetition is allowed, that is 2 choices
There is no bucket 3 & 4, as we have to find 2 letters word such as CC, AA, CA, CD, RD, RA etc
If you multiple all these choices: 4 * 4, you will get the number of possible 2 letter words with“ C A R D” with repetition.
Thus number of possible 2 letter words are: 4 * 4 = 16 words when repetition is allowed.
Thus we have seen 4 scenarios
1. Find number of arrangement of compete set objects without repetition.
1. Find number of arrangement of compete set objects with repetition.
2. Find number of arrangement using smaller set objects without repetition.
3. Find number of arrangement using smaller set objects with repetition.
Lets create general formula for these scenarios.
Note that the scenario we have discussed has all the elements unique { C, A, R, D }, thus we can call it set.
Case 1: Without repetition & Permutation of whole set Permutation = n *( n-1) *( n-2)… 1 Case 2: With repetition & Permutation of whole set Permutation = n * n * n … n times