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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: 57022
A certain university will select 1 of 7 candidates eligible  [#permalink]

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11 Mar 2014, 03:27
4
15
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Difficulty:

5% (low)

Question Stats:

85% (01:41) correct 15% (02:09) wrong based on 638 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, 03:27
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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, 04:19
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, 07:27
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, 05:22
identical positions should be divided by 2 such as 7*10*9/1*2=315
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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, 07:17
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, 12:23
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, 12:02
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  [#permalink]

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04 Aug 2019, 13:33
ScottTargetTestPrep wrote:
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.

Hi,
Can this be done by adding the total candidates as well as the position? I.e. 17 candidates and 3 positions will give 17C3?

Just curious

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

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05 Aug 2019, 08:51
Shef08 wrote:
ScottTargetTestPrep wrote:
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.

Hi,
Can this be done by adding the total candidates as well as the position? I.e. 17 candidates and 3 positions will give 17C3?

Just curious

Posted from my mobile device

No because we are filling positions in "specific departments".
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# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: A certain university will select 1 of 7 candidates eligible   [#permalink] 05 Aug 2019, 08:51
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