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M18-23

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M18-23  [#permalink]

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New post 16 Sep 2014, 01:04
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

91% (00:56) correct 9% (01:50) wrong based on 177 sessions

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Re M18-23  [#permalink]

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New post 16 Sep 2014, 01:04
1
Official Solution:

10 engineers and 3 designers have volunteered to work on a project. If the project team must consist of 3 engineers and 2 designers, in how many different ways can the team be formed?

A. 360
B. 400
C. 420
D. 450
E. 600

Engineers can be selected in \(C_{10}^{3} = \frac{10!}{7! 3!} = 120\) different ways. Designers can be selected in \(C_{3}^{2} = \frac{3!}{2! 1!} = 3\) different ways. The answer to the question is therefore \(120*3 = 360.\)

Answer: A
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Re: M18-23  [#permalink]

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New post 30 Jun 2017, 22:55
I agree with the explanation.

One of the important rule learnt is AND means multiplication and OR means addition.

This is not a probability question.

It is asking ways to arrange the engineers and designers and the order doesn't matter. A simple anagram question.
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Re: M18-23  [#permalink]

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New post 04 Jul 2017, 05:32
I had a simple question, the problem says how many different ways, order doesn't matter, shouldn't we be using permutation than combination ? I think it should be 10P3 x 3P2
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Re: M18-23  [#permalink]

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New post 04 Jul 2017, 05:39
Ganeshsrinivasan wrote:
I had a simple question, the problem says how many different ways, order doesn't matter, shouldn't we be using permutation than combination ? I think it should be 10P3 x 3P2


Yes, order does not matter and that's why we are using combinations and not permutation. 10C3 gives different unordered groups of 3 out of 10 and 3C2 gives different unordered groups of 2 out of 3. Multiplying those two gives the answer.
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Re: M18-23   [#permalink] 04 Jul 2017, 05:39
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