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# What is the number of different ways to choose a chairman two deputies

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Senior Manager
Joined: 24 Jul 2009
Posts: 259
What is the number of different ways to choose a chairman two deputies [#permalink]

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28 May 2010, 14:09
1
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Difficulty:

5% (low)

Question Stats:

85% (00:54) correct 15% (00:47) wrong based on 43 sessions

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What is the number of different ways to choose a chairman, two deputies, and two assistants for the class committee out of 7 students up for elections.

A. $$7C1*6C1*5C1*4C1*3C1$$
B. $$7C1*6C2*4C2$$

Which solution is correct ? And Why ?
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 44655
Re: What is the number of different ways to choose a chairman two deputies [#permalink]

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28 May 2010, 15:09
nverma wrote:
What is the number of different ways to choose a chairman, two deputies, and two assistants for the class committee out of 7 students up for elections.

A. $$7C1*6C1*5C1*4C1*3C1$$
B. $$7C1*6C2*4C2$$

Which solution is correct ? And Why ?

Correct answer is B: $$C^1_7*C^2_6*C^2_4=630$$.

$$C^1_7$$ - # of ways to choose 1 chairmen out of 7;
$$C^2_6$$ - # of ways to choose 2 deputies out of 6 members left;
$$C^2_4$$ - # of ways to choose 2 assistants out of 4 members left.

Answer A is not correct because it's counting # of different ways to choose 2 deputies out of 6 as $$C^1_6*C^1_5$$ (and next # of ways to choose 2 assistants out of 4 as $$C^1_4*C^1_3$$) which is not right. $$C^1_6*C^1_5$$ will have duplications in it and needs to be divided by 2! (# of peoples), which then gives the same answer as $$C^2_6$$.

Consider this: in how many different ways we can choose 2 different letters out of A, B, and C?

AB
AC
BC

Only 3, which can be obtained by $$C^2_3=3$$, another way would give incorrect answer - $$C^1_3*C^1_2=6$$.

The original question can be solved in another way:
Members: 1 - 2 - 3 - 4 - 5 - 6 - 7. Positions: C (chairmen), D (deputy), D (deputy), A (assistant), A (assistant), N (no position), N (no position): CDDAANN. # of ways to assign each letter (each position) to the members would be the # of permutations of 7 letters CDDAANN = $$\frac{7!}{2!2!2!}=630$$.

1-2-3-4-5-6-7
C-D-D-A-A-N-N
D-C-D-A-A-N-N
D-D-C-A-A-N-N
...
...

Hope it helps.
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Re: What is the number of different ways to choose a chairman two deputies [#permalink]

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13 Dec 2017, 00:32
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Re: What is the number of different ways to choose a chairman two deputies   [#permalink] 13 Dec 2017, 00:32
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# What is the number of different ways to choose a chairman two deputies

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