Problem Solving

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

crazy4gmat 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

Mathematics = 7C1 = 7

Science = 10C2 = 45

total combinations = 7*45 = 315.

OA?

OA?

the question seems to be flawed... I looks like they wanted us to think there was a pool of 17 candidates

luisengard wrote:OA?

the question seems to be flawed... I looks like they wanted us to think there was a pool of 17 candidates

Bustards!

Can someone clearly explain why its 7C1 * 10C2 and not 7C1 plus 10C2? Am missing some core concept here

gokyada wrote:Can someone clearly explain why its 7C1 * 10C2 and not 7C1 plus 10C2? Am missing some core concept here

Hi,

here's the rule known as the Fundamental Counting Principle:

If there are m ways to make one selection, and n ways to make another selection, then as long as those selections are independent, there are m*n ways to make both selections.

Here's another way to remember it:

whenever you make MULTIPLE selections, MULTIPLY; and

whenever you make ALTERNATIVE selections, ADD.

Since we're selecting one person for the first position AND 2 people for the second pair of positions, we MULTIPLY.

Words such as "and", "both" and "all" indicate multiple selections; words such as "or", "at least" and "at most" indicate alternative selections.

Here's a simple example to help us understand the rule.

A class contains 3 girls and 2 boys. If one girl and one boy are going to be selected to dance together, how many different pairs of dancers can be formed?

Let's call the girls A, B and C and the boys F and G. We're choosing 1 out of 3 girls, so there are 3 different selections we can make; we're choosing 1 out of 2 boys, so there are 2 different selections we can make. We're choosing a girl AND a boy, so we multiply:

3*2 = 6

If we were to actually count the pairs, we'd confirm our calculation:

AF
AG
BF
BG
CF
CG

As you can see from the list, for each girl we select, there are two possible boys - that's why we multiply.

Now let's change the question a bit:

A class contains 3 girls and 2 boys. If one girl or one boy is going to be chosen to be class president, how many different presidents can be chosen?

Keeping the same names, we see that out of the girls, we have 3 different choices; out of the boys, we have 2 different choices. Since we're selecting 1 girl OR 1 boy, we ADD the results: 3+2=5.

Again, if we were to list them out, we confirm our math:

A
B
C
D
E

As you can see from this list, there's no combining at all, so we simply add the results together.

Going back to the original question, we have 7 possible choices for the first position and 45 for the second pair; since we're choosing 1 of 7 AND 2 of 10, we multiply to get the final answer. If we were to actually list it out (calling the first group A through F and the second group G through P):

AGH
AGI
AGJ
.
.
.

we quickly see that there are a LOT more than 17 possibilities.

because its one of those AND cases where both events should occur together two events being
1. 1 selected for maths dept out of 7
2. 2 candidates selected for comp sc dept out of 10
we will multiply both probabilities.

when we have the OR scenario (either selecting one for maths dept OR 2 for comp sc dept), in that case we will add the 2 probabilities to get the total probability.

gokyada wrote:Can someone clearly explain why its 7C1 * 10C2 and not 7C1 plus 10C2? Am missing some core concept here

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 have done this right, but i have a question yet, if i use one of the candidates, instead none, is eligible for a position in both department then what will be the process to solve this.

Thank you...

I have done this right, but i have a question yet, if i use one of the candidates, instead none, is eligible for a position in both department then what will be the process to solve this.

Thank you...

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 one of the candidates is eligible for both departments but may be selected to fill at most one of the available positions, how many different sets of 3 candidates are there to fill the 3 positions?

A. 42
B. 70
C. 140
D. 165
E. 315

Let A = the candidate eligible for both departments.

Case 1: A is selected for the math position
From the remaining 9 candidates eligible for the computer science department, the number of ways to select 2 = 9C2 = (9*8)/(2*1) = 36.

Case 2: A is NOT selected for the math position
Number of remaining options for the math position = 6. (Any of the 6 other candidates.)
From the 10 candidates eligible for the computer science department, the number of ways to select 2 = 10C2 = (10*9)/(2*1) = 45.
To combine these options, we multiply:
6*45 = 270.

Total options = 36 + 270 = 306.

There are two combinations here. One is that of maths and the other of computer science. Now in my opinion these two combinations are to be multiplied because of a core principle i.e. AND=multiply & OR = Add.

Now for the maths committee, we have to choose 1 member out of a possible 7. So that gives me a 7C1!*6! = 7 combinations. Because for every 1 possible candidate there are 6 candidates who will be not selected. Hence the denominator is 1!*6!.

Similarly for the computer science committee, there 10C2!*8! = 45 combinations.

So total no of combinations =45*7 = 315
So the answer is E.

Then again this is what my limited thinking has come up with and i am sure that I am wrong because I suck at probability or combinations.

However I do have a question. This seems like a combination problem then why is it tagged as Probability ?