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A certain university will select 1 of 7 candidates eligible [#permalink]
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20 Oct 2010, 09:11
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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 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. Thanks a lot in advance for your help.
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).
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)
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.
A certain university will select 1 of 7 candidates eligible [#permalink]
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15 Jan 2015, 14:51
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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.
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. Thanks a lot in advance for your help.
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).
Answer: E.
hey Bunuel sorry for off topic, but how would it look like without the restriction ??
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.
Answer: E
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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. Thanks a lot in advance for your help.
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).
Answer: E.
Hi Bunuel,
What would be the process if the restriction was not there? Bunuel
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
Answer: D
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
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21 Apr 2018, 07:54