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# For an international Mathematics Olympiad, six delegates

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Intern
Joined: 07 May 2014
Posts: 19

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02 Jun 2014, 22:50
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Difficulty:

45% (medium)

Question Stats:

69% (02:25) correct 31% (02:53) wrong based on 132 sessions

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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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03 Jun 2014, 00:16
1
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 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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Joined: 02 Sep 2009
Posts: 58340

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03 Jun 2014, 01:25
1
1
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).

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15 Feb 2019, 18:48
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.

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