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Committee X has 4 members, committee Y has 5 members, and these commit

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janchang
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Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   02 Sep 2018, 12:16
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Committee X has 4 members, committee Y has 5 members, and these committees have no members in common. If a task force is to be formed consisting of one member of X and one member of Y, how many different task forces are possible?

A) 6
B) 9
C) 10
D) 20
E) 36
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D
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fskilnik
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   02 Sep 2018, 13:21
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janchang wrote:
Committee X has 4 members, committee Y has 5 members, and these committees have no members in common. If a task force is to be formed consisting of one member of X and one member of Y, how many different task forces are possible?

A) 6
B) 9
C) 10
D) 20
E) 36


Immediate application of the Multiplicative Principle:

If you may choose the member of X among 4 possibilities, and each possibility allows you to choose the member of Y among 5 possibilities, there are 4*5 = 20 possibilities for the task force!

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fskilnik.
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   02 Sep 2018, 13:47
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Solution



Given:
    • Committee X has 4 members
    • Committee Y has 5 members
    • A task force is formed consisting of one member of X and one member of Y

To find:
    • The number of task force possible.

Approach and Working:

    • The number of task force possible= Number of ways to select 1 member from committee X * Number of ways to select 1 member from committee Y
      o Hence, total task force possible= 4*5= 20

Hence, the correct answer is option D.

Answer: D
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   10 Sep 2020, 19:54
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This is a question of choices and hence is a question on combinations. Order does not matter.

\(^nC_r = \frac{n!}{(n-r)!*r!}\)

Important values to remember are that \(^nC_1 = n\), \(^nC_n = 1\) and \(^nC_0 = 1\)


Committee X has 4 members: Choosing one out of 4 = \(^4C_1 = 4\)

Committee Y has 5 members: Choosing one out of 5 = \(^5C_1 = 5\)


Since we need to ensure that one of each has to be chosen to create a task force, the even is not completed until one of each is chosen. Therefore we multiply the 2 values. The total possibilities = 4 * 5 = 20


Option D

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100mitra
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   18 Sep 2021, 06:40
Correct option D - 20 [4C1 x 5C1 = 4x5 = 20]
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   19 May 2023, 09:59
Isn't there double counting happening here? i.e. we are counting picking member 1 from committee x and member 1 from committee y, and then again we are counting picking member 1 from committee y and member 1 from committee x?
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink] New post   05 Oct 2023, 10:04
kelly_jacques wrote:
Isn't there double counting happening here? i.e. we are counting picking member 1 from committee x and member 1 from committee y, and then again we are counting picking member 1 from committee y and member 1 from committee x?


Hi! Actually, no. Please, read the best explanation above, which provides the number of combinations. This means that possible repetitions are excluded.
Also, we use the 4 times 5 because it is a condition to find a max possible variety of 2 combinations at once (using AND as a multiplicator).
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Re: Committee X has 4 members, committee Y has 5 members, and these commit [#permalink]
05 Oct 2023, 10:04
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