A certain university will select 1 of 7 candidates eligible to fill a

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

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

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!

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

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

Hi ScottTargetTestPrep
Could you help me with my query?
What role does this phrase play in the question? Does it affect our calculation at all?

Thank you,
Dablu

The fact that none of the candidates is eligible for both departments play a significant role in the question. To compare, suppose that there was one person who is eligible for a position in both departments. First of all, since that one person is included both in the 7 candidates for the math department and 10 candidates for the CS department, there are no longer 17 total candidates; there are only 7 + 10 - 1 = 16 total candidates. Further, the number of sets of 3 candidates change as well. For the math position, we still have 7 choices; however, the number of ways we can fill the CS position depends on whether the person filling the math position is the one eligible for both positions or not. If it is, then there are only 9 people you can choose from to fill the CS position; so in this scenario, the number of ways to choose 2 people from a total of 9 people is 9C2 = 9!/(2!*7!) = (9*8)/2 = 36. On the other hand, if the person chosen for the math department is one of the remaining six people, then we can fill the CS positions by choosing 2 people from a total of 10 people, which can be done in 10C2 = 45 ways, as calculated before. Thus, when there's one person eligible for both positions, the number of ways to fill the three positions is 36 + 6*45 = 306.

If you're curios what happened to the 315 - 306 = 9 ways to fill the three positions, let X be the person eligible for both positions. Since X is included in both the 7 and 10 people, the choices X + XA, X + XB, ... are no longer possible, where A, B, ... denote the people eligible for the CS department besides X. As there are 9 people eligible for the CS department besides X, the total number of ways is reduced by 9.

Hi ScottTargetTestPrep Thank you for replying great explanation!



I totally got this approach: The total no of possible combinations - 9 combinations with are not possible

Sorry for bothering again, could tell me how did you deduce this?

When tried solving it directly I got this:

let X be the person eligible for both positions.

1st case: If X is selected in Math department
7C1 and 9C2
or
2nd case: If X is selected in CS department.
10C2 and 6C1

= 7C1 * 9C2 + 10C2 * 6C1 = 252 + 270 = 522. Where am I going wrong?

Thank you,
Dablu

The only mistake in your calculations is that when X is selected for the position in the Math department, there are not 7C1 = 7 choices; since we are considering the case when X is already selected, the number of ways to fill the position in the Math department is 1C1 = 1. Thus, it should have been 1C1 * 9C2 + 6C1 * 10C2 = 1 * 36 + 6 * 45 = 36 + 270 = 306.

The question involves selection i.e. combinations.
Select 1 out of 7 people in 7C1 = 7 ways.
Select 2 out of 10 people in 10C2 = 10*9/2 = 45 ways

So we have total 7 * 45 = 315 ways of selecting 3 people.

Answer (E)

Check out a discussion on Combinations here: https://youtu.be/tUPJhcUxllQ