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# A certain university will select 1 of 7 candidates eligible

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

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17 Nov 2007, 05:35
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-university-will-select-1-of-7-candidates-eligible-103273.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Dec 2013, 06:20, edited 1 time in total.
Edited the question and added the OA.
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17 Nov 2007, 06:37
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1C7*2C10 = 7*45 = 315

the answer is (E)

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17 Nov 2007, 06:58
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Ferihere wrote:
KillerSquirrel wrote:
1C7*2C10 = 7*45 = 315

the answer is (E)

the explanation would be greatly appreciated...

To find the ways to choose one item out of a group of items we can use the combinations formula (i.e xCn = n!/((n-x)!*x!)).

So the ways to choose one out of seven is 1C7 = 7!/6!*1! = 7 and two out of ten is 10!/8!*2! = 45.

Total ways for both are ---> 7*45 = 315

Alternatively you can say that:

1/7*2/10*1/9 = 2/630 = 1/315

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Re: Set #1 (probability question) [#permalink]

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24 Jul 2009, 18:10
7C1X10C2
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Re: Set #1 (probability question) [#permalink]

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27 Sep 2009, 01:22
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

Soln:
1 out of 7 candidates can be chosen in 7 ways for mathematics department

2 out of 10 candidates can be chosen in 10C2 ways to fill two identical positions in Comp Sci department

Thus total number of ways = 7 * 10C2 = 315
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Re: Set #1 (probability question) [#permalink]

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14 Feb 2010, 08:24
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Ferihere wrote:
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

thank you in advance

Maths Dept = 7c1 = 7

CS dept = 10c2 = 45

Therefore no of combinations = 7 x 45 = 315 ... E
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Re: Set #1 (probability question) [#permalink]

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08 Sep 2010, 02:37
damn i think too much...was making a mess of this problem
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Re: Set #1 (probability question) [#permalink]

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18 Jun 2011, 05:48
agree with 315.. 7c1 * 10c2
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Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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05 Dec 2011, 10:26
(7c1*10c1*9c1)/2! = 315
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Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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06 Dec 2011, 18:56
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There are a few possible areas in this problem where you can go wrong.

First off: does the solution require combinations or permutations?

For the mathematics dept., in which you are selecting 1 person from 7, it is irrelevant whether you use combinations or permutations – the answer is the same. Also, whenever you see nC1, remember that the answer is n (don’t feel you have to set up all the factorials).

With the computer science dept., you have two identical positions. Now you have to address the initial question: combinations or permutations. The order in which any two candidates are chosen (say, candidate A and candidate B) is irrelevant (AB is the same as BA) thus you should use the combinations formula. 10C2.

The quick math in this case is as follows: when you have nC2, where n is any integer greater than or equal to 4, multiply n(n-1)/2 to get the answer. In this case n = 10 so (10)(9)/2 = 45.

The second trouble spot is whether to add or multiple the 45 and the 7. Because each of the 7 math departments can be matched up with any 45 of the comp. sci. dept., you want to multiply. The 7 different possibilities for group A can be matched up with the 45 different possibilities from Group B to get: 7 x 45 = 315.

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

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07 Dec 2013, 11:18
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Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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08 Dec 2013, 06:21
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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).

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-university-will-select-1-of-7-candidates-eligible-103273.html
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Re: A certain university will select 1 of 7 candidates eligible   [#permalink] 08 Dec 2013, 06:21
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