krishnasty wrote:

How many ways can a selection be done of 5 letters out of 5 A's, 4B's, 3C's, 2D's and 1 E.

A. 60

B. 75

C. 71

D. 121

E. 221

Notice that you won't see such question on the GMAT. So, just for fun.

We have the following letters: {AAAAA}, {BBBB}, {CCC}, {DD}, {E}

There are 7 different cases of 5 letter selections possible:

(5) - all letters are alike - 1 way, all A's;

(4, 1) - 4 letters are alike and 1 different - \(C^1_2*C^1_4=8\), where \(C^1_2\) is # of ways to choose which letter provides us with 4 letters from 2 (A or B) and \(C^1_4\) is # of ways to choose 5th letter from 4 letters left;

(3, 2) - 3 letters are alike and other 2 are also alike - \(C^1_3*C^1_3=9\), where \(C^1_3\) is # of ways to choose which letter provides us with 3 letters from 3 (A, B or C) and \(C^1_3\) is # of ways to choose which letter provides us with 2 letters from 3 (for example if we choose A for 3 letters then we can choose from B, C or D for 2 letters);

(3, 1, 1) - 3 letters are alike and other 2 are different - \(C^1_3*C^2_4=18\), where \(C^1_3\) is # of ways to choose which letter provides us with 3 letters from 3 (A, B or C) and \(C^2_4\) is # of ways to choose which 2 letters provides us with one letter each;

(2, 2, 1) - 2 letters are alike, another 2 letters are also alike and 1 is different - \(C^2_4*C^1_3=18\), where \(C^2_4\) is # of ways to choose which 2 letters provides us with 2 letters from 4 (A, B, C or D) and \(C^1_3\) is # of ways to choose which provides us with 5th letter from 3 letters left;

(2, 1, 1, 1) - 2 letters are alike and other 3 are different - - \(C^1_4*C^3_4=16\), where \(C^1_4\) is # of ways to choose which letter provides us with 2 letters from 4 (A, B, C or D) and \(C^3_4\) is # of ways to choose which 3 letters out of 4 provides us with one letter each;

(1, 1, 1, 1, 1) - all letters are distinct - 1 way (A, B, C, D, E).

Total: 1+8+9+18+18+16+1=71.

Answer: C.

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