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