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# A certain university will select 1 of 7 candidates eligible

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Math Expert
Joined: 02 Sep 2009
Posts: 51121
A certain university will select 1 of 7 candidates eligible  [#permalink]

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11 Mar 2014, 02:27
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Difficulty:

5% (low)

Question Stats:

85% (01:42) correct 15% (02:09) wrong based on 687 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

(A) 42
(B) 70
(C) 140
(D) 165
(E) 315

Problem Solving
Question: 151
Category: Arithmetic Elementary combinatorics
Page: 82
Difficulty: 600

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Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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11 Mar 2014, 02:27
SOLUTION

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

(A) 42
(B) 70
(C) 140
(D) 165
(E) 315

As "none of the candidates is eligible for a position in both departments" then we have 7+10=17 candidates.

$$C^1_7*C^2_{10}=7*45=315$$: $$C^1_7$$ - choosing 1 from 7 and $$C^2_{10}$$ choosing 2 from 10 when order doesn't matter as 2 positions in computer science department are identical (XY is the same as YX).

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Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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11 Mar 2014, 03:19
4
A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

(A) 42
(B) 70
(C) 140
(D) 165
(E) 315

Sol: 1 out of 7 candidates will be selected in 7!/6! or 7 ways
2 out of 10 candidates can be selected by 10!/8!*2! or 45 ways

No. of ways in which 3 different sets can be filled is =45*7 =315
Ans is E
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Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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11 Mar 2014, 06:27
1
Bunuel wrote:
RESERVED FOR A SOLUTION.

7c1=7

10c2=45

7*45= 315
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Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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08 Jul 2017, 04:22
identical positions should be divided by 2 such as 7*10*9/1*2=315
Intern
Joined: 23 May 2015
Posts: 24
Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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22 Jun 2018, 06:17
Bunuel wrote:
SOLUTION

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

(A) 42
(B) 70
(C) 140
(D) 165
(E) 315

As "none of the candidates is eligible for a position in both departments" then we have 7+10=17 candidates.

$$C^1_7*C^2_{10}=7*45=315$$: $$C^1_7$$ - choosing 1 from 7 and $$C^2_{10}$$ choosing 2 from 10 when order doesn't matter as 2 positions in computer science department are identical (XY is the same as YX).

Bunuel What is the purpose of the statement " If none of the candidates is eligible for a position in both departments" in the question. Does it have any significance?
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Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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22 Jun 2018, 11:23
So basically, in the first set we have to choose 1 individual out of the 7 in which case the total number of possible choices would be obviously 7.

In the second set we have to choose a combination of 2 individuals out of the 10. I found it easier to visualize this like this:
Possible Individuals --> A B C D E F G H I J --> Each letter represents 1 specific individual
Combinations with Person A --> AB AC AD AE AF AG AH AI AJ = 9 unique possible combinations
Combinations with Person B --> BC BC BE BF BG BH BI BJ = 8 unique possible combinations
Combinations with Person C --> CD CE CF CG CH CI CJ = 7 unique possible combinations
and so on...

What the second set will then look like once we have all the possible unique combinations is basically:
9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45

Now we just multiply the number of possible choices from set 1 with the unique number of combinations from set 2 and we get:
7 x 45 = 315

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Re: A certain university will select 1 of 7 candidates eligible  [#permalink]

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25 Jun 2018, 11:02
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

(A) 42
(B) 70
(C) 140
(D) 165
(E) 315

1 of 7 candidates can fill the math position in 7C1 = 7 ways.

2 of 10 candidates can fill the two computer science positions in 10C2 = 10!/(2! x 8!) = (10 x 9)/2! = 45 ways.

Thus, the total number of ways to fill the three positions is 7 x 45 = 315.

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Re: A certain university will select 1 of 7 candidates eligible &nbs [#permalink] 25 Jun 2018, 11:02
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