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There are X people at a dance, where X is an integer greater than 1.

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There are X people at a dance, where X is an integer greater than 1.  [#permalink]

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New post 26 Jun 2018, 05:16
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  45% (medium)

Question Stats:

65% (01:40) correct 35% (01:29) wrong based on 57 sessions

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There are X people at a dance, where X is an integer greater than 1. The number of different pairs of people who can be partnered up for the waltz is Z. Which of the following could be the value of Z ?

A. 2

B. 6

C. 9

D. 14

E. 20

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Re: There are X people at a dance, where X is an integer greater than 1.  [#permalink]

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New post 26 Jun 2018, 06:48
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Solution



Given:
    • There are X people at a dance, where X is an integer greater than 1
    • The number of different pairs of people who can be partnered up for the waltz is Z

To find:
    • Out of the given options, which one can be a possible value of Z

Approach and Working:
Out of X people, if a pair of 2 is to be formed, that can be done in XC2 ways
    • Therefore, Z = \(^XC_2\)

It is given that X is an integer and X > 1

Trying with different possible values of X, we get
    • \(^2C_2 = 1\) [not given in the options]
    • \(^3C_2 = 3\) [not given in the options]
    • \(^4C_2 = 6\) [given in the options]

Hence, the correct answer is option B.

Answer: B
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Re: There are X people at a dance, where X is an integer greater than 1.  [#permalink]

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New post 28 Jun 2018, 12:25
Bunuel wrote:
There are X people at a dance, where X is an integer greater than 1. The number of different pairs of people who can be partnered up for the waltz is Z. Which of the following could be the value of Z ?

A. 2

B. 6

C. 9

D. 14

E. 20


The # of pairs that can be chosen out of X people is Xc2 = X!/(2!*(X-2)! = X(X-1)/2

Hence we have Z = X(X-1)/2 or 2Z = X(X-1), a product of two consecutive integers

Lets check the answer options, Z = 2, 2Z = 4 = 2*2 or 1*4....Not possible

Z = 6, 2Z = 12 = 4*3....Possible

Hence Z = 6

Answer B.


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GyM
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Re: There are X people at a dance, where X is an integer greater than 1. &nbs [#permalink] 28 Jun 2018, 12:25
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