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For an international Mathematics Olympiad, six delegates

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dxx
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For an international Mathematics Olympiad, six delegates [#permalink] New post   02 Jun 2014, 22:50
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For an international Mathematics Olympiad, country D will send six delegates in total — two will be supervisors and four will be contestants. There are 210 ways in which the six delegates can be chosen and there are seven candidates competing for the four contestants’ places available. How many candidates are competing for the two supervisors’ slots available?

A. 1
B. 2
C. 3
D. 4
E. 5
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D
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Re: For an international Mathematics Olympiad, six delegates [#permalink] New post   03 Jun 2014, 00:16
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dhirajx wrote:
For an international Mathematics Olympiad, country D will send six delegates in total — two will be supervisors and four will be contestants. There are 210 ways in which the six delegates can be chosen and there are seven candidates competing for the four contestants’ places available. How many candidates are competing for the two supervisors’ slots available?

A - 1
B - 2
C - 3
D - 4
E - 5


We can solve this question by using options:

Option A wrong as 2 supervisor need to be selected.
Going by statement: 4 contestent choosen from 7 -- 7C4 == 35
and total ways =210
Consider 2 supervisor will be choosen from : X people.
So as per question: 35*X= 210
X= 6
So by using options,
Option A already decided wrong.
Option B 2C2 will be 1 not 6.
Option C 3C2 will be 3 not 6
Option D 4C2 will be 6 correct answer
Option E 5C2 will be 10 not 6.
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Re: For an international Mathematics Olympiad, six delegates [#permalink] New post   03 Jun 2014, 01:25
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dhirajx wrote:
For an international Mathematics Olympiad, country D will send six delegates in total — two will be supervisors and four will be contestants. There are 210 ways in which the six delegates can be chosen and there are seven candidates competing for the four contestants’ places available. How many candidates are competing for the two supervisors’ slots available?

A. 1
B. 2
C. 3
D. 4
E. 5


2 supervisors out of x;
4 contestants out of 7.

\(C^2_x*C^4_7=210\) --> \(\frac{x!}{2!(x-2)!}*35=210\) -->\(\frac{x(x-1)}{2}*35=210\) --> \(x(x-1)=12\) --> \(x=4\) (discard the negative root).

Answer: D.
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Re: For an international Mathematics Olympiad, six delegates [#permalink] New post   15 Feb 2019, 18:48
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dxx wrote:
For an international Mathematics Olympiad, country D will send six delegates in total — two will be supervisors and four will be contestants. There are 210 ways in which the six delegates can be chosen and there are seven candidates competing for the four contestants’ places available. How many candidates are competing for the two supervisors’ slots available?

A. 1
B. 2
C. 3
D. 4
E. 5


We can let n = the number of candidates vying for the 2 supervisors’ slots and create the equation:

nC2 x 7C4 = 210

[Note: nC2 = n!/[(n - 2)! x 2!] =[ n x (n-1) x (n-2) x (n-1) x … x 1] / {[(n-2) x (n-1) x … x 1] x 2!}. We see that all the factors in the numerator cancel with those in the denominator, except n x (n-1). Thus, we have nC2 = n(n - 1)/2]

n(n - 1)/2 x (7 x 6 x 5 x 4)/(4 x 3 x 2) = 210

(n^2 - n)/2 x 35 = 210

(n^2 - n)/2 = 6

n^2 - n = 12

n^2 - n - 12 = 0

(n - 4)(n + 3) = 0

n = 4 or n = -3

Since n can’t be negative, then n must be 4.

Answer: D
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Re: For an international Mathematics Olympiad, six delegates [#permalink]
15 Feb 2019, 18:48
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