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# How many ways can a selection be done of 5 letters out of 5

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How many ways can a selection be done of 5 letters out of 5 [#permalink]

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11 Jul 2011, 23:21
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(N/A)

Question Stats:

22% (00:00) correct 78% (01:59) wrong based on 9 sessions

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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Feb 2012, 05:20, edited 1 time in total.
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Re: selection be done of 5 letters [#permalink]

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07 Feb 2012, 05:19
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.

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Re: How many ways can a selection be done of 5 letters out of 5 [#permalink]

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07 Feb 2012, 06:50
Even though it would not appear on GMAT, it is still a good questions to tackle. i am always weak with combinations.
Re: How many ways can a selection be done of 5 letters out of 5   [#permalink] 07 Feb 2012, 06:50
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