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# A certain university will select 1 of 7 candidates eligible

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Manager
Joined: 19 Sep 2010
Posts: 54
Location: Pune, India
A certain university will select 1 of 7 candidates eligible [#permalink]

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20 Oct 2010, 09:11
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5% (low)

Question Stats:

85% (01:18) correct 15% (01:32) wrong based on 275 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 45222

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20 Oct 2010, 09:16
Expert's post
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SoniaSaini 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

I could solve it half.like 7C1. but here two positions are identical. after that I'm getting lost. Please help me come out form there.

Sonia

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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Joined: 13 Jan 2012
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23 Feb 2012, 13:52
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Math 1 of 7 ==> 7 possibilities
CSC 2 of 10 ==> 10C2 possibilities = 45 (not 10P2 because the positions are identical, implying their order does not matter. When order does not matter, we use the C formula, instead of P)

Total: 7 * 45 = 315
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Joined: 27 Nov 2011
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28 Feb 2012, 02:03
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I am like those, who find Permutation, combination and probablity a nightmare. However, my short mantra do deal with the easy set is :-

Whenever you need to choose or select, use combination

eg:- choose/select r from n is:- nCr

Whenever you need to choose and order, use combination formula and multiply by with the number you need to order.

eg:-Choose/select r from n is:- nCr, followed by order of r is nCr*r!
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Joined: 02 Sep 2009
Posts: 45222

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28 Feb 2012, 02:16
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Avijit wrote:
I am like those, who find Permutation, combination and probablity a nightmare. However, my short mantra do deal with the easy set is :-

Whenever you need to choose or select, use combination

eg:- choose/select r from n is:- nCr

Whenever you need to choose and order, use combination formula and multiply by with the number you need to order.

eg:-Choose/select r from n is:- nCr, followed by order of r is nCr*r!

Generally:
The words "Permutation" and "Arrangement" are synonymous and can be used interchangeably (order matters).
The words "Combination" and "Selection" are synonymous and can be used interchangeably.

Check Combinatorics and Probability chapters of Math Book to foind out more on these topics:
math-combinatorics-87345.html
math-probability-87244.html

Combinations questions to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52

Probability questions to practice:
DS: search.php?search_id=tag&tag_id=33
PS: search.php?search_id=tag&tag_id=54

Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html

Hope it helps.
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Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 423
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
A certain university will select 1 of 7 candidates eligible [#permalink]

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15 Jan 2015, 14:51
1
KUDOS
For me, I cannot say that I understand the difference, but somehow I manage to solve some of these problems. This one was very easy for me, thankfully!

So, I cannot understand the Choose method. What I did was this:

1 of 7 will be chosen for the math
2 of 10 will be chosen for the computer
None of the 3 chosen people can be in more thn one deparments.

We can choose any of the 7 candidates for the math dep., which gives as 7 selections.
We can choose 2 of the 10 candidates for the computer dep., which gives us 2 selections and 8 rejections.
So, the way to find how many different selections of 2 candidates we can have for the computer dep., we do:
10! / 2!*8! = (9*10) / 2 = 90 / 2 = 45.

We are multiplying our individual selections: 7*45 = 315

In the bolded part, we don't have to multiply all of the numbers, as those in 8! are included in 10!, so we simplify instead.
Manager
Joined: 26 Mar 2017
Posts: 149
Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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13 May 2017, 10:32
Bunuel wrote:
SoniaSaini 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

I could solve it half.like 7C1. but here two positions are identical. after that I'm getting lost. Please help me come out form there.

Sonia

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).

hey Bunuel sorry for off topic, but how would it look like without the restriction ??

is 7*10*9 correct ?
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Posts: 297
Location: India
Schools: IIMB
GMAT 1: 550 Q42 V28
GPA: 3.96
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Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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18 May 2017, 23:51
here in the science order does not matter so the possible combination will be 7*10*9 now divide by 2 = 315
hence E
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Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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22 May 2017, 18:57
SoniaSaini 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

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

2 of 10 candidates can fill the computer science position in 10C2 = (10 x9)/2! = 45 ways.

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

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Intern
Joined: 22 Nov 2017
Posts: 9
Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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26 Jan 2018, 14:33
Bunuel wrote:
SoniaSaini 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

I could solve it half.like 7C1. but here two positions are identical. after that I'm getting lost. Please help me come out form there.

Sonia

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).

Hi Bunuel,

What would be the process if the restriction was not there? Bunuel

Thanks,
SVP
Joined: 12 Sep 2015
Posts: 2455
Re: A certain university will select 1 of 7 candidates eligible [#permalink]

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21 Apr 2018, 07:54
Expert's post
Top Contributor
SoniaSaini 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

One approach is to apply counting methods:

P(exactly 2 women) = [# of teams with exactly 2 women] / [total # of teams possible]

# of teams with exactly 2 women
Take the task of selecting 2 women and 2 men and break it into stages.
Stage 1: Select 2 women for the team. There are 5 women to choose from, so this can be accomplished in 5C2 ways.
Stage 2: Select 2 men for the team. There are 3 men to choose from, so this can be accomplished in 3C2 ways.
By the Fundamental Counting Principle (FCP), the total number of teams with exactly 2 women = (5C2)(3C2) = (10)(3) = 30

# of teams possible
There are 8 people altogether and we must choose 4 of them.
This can be accomplished in 8C4 ways, which equals 70 ways

P(exactly 2 women) = 30/70
= 3/7

ASIDE: To learn how to calculate combinations (like 5C2) in your head, watch this video:

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Re: A certain university will select 1 of 7 candidates eligible   [#permalink] 21 Apr 2018, 07:54
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