Question

There are 8 people interested in serving on a selection committee for a new faculty member. The Dean will select 3 to serve. How many ways might she select the three member team?

Answer 1

There are 8 people to choose from for the first person. The second person, there will be 7 different people left, and for the last person, only 6 people. The order that you choose them does not matter. The answer is 8*7*6. If you imagine the people as a,b,c,d,e,f,g, and h, think about the choices you have left. If you choose person a first, there are 7 different combinations for choosing the second person. a,b; a,c; a,d; a,e; a,f; a,g; a,h. So there are 8*7 combinations for the first two people. Then choosing the last, there are only 6 available to chose from.

Answer 2

thank you!!!

Answer 3

how many four letter combinations of math can you make if you use each letter only once?
would 4*3*2*1 be my answer?

Answer 4

Is each letter chosen from the set of 26?

Answer 5

no, just from the word MATH

Answer 6

So that is a permutation (not combination) because order counts.
Yes, in the first position there are four choices, then three, then two, then one, which makes 4! permutations.

Answer 7

okay, thanks again for the help !

Answer 8

In the previous problem, order does not matter, so for three members in the committee, we have overcounted 3! times (ABC,ACB,BAC,BCA,CAB,CBA), so the result of 8*7*6 must be divided by 3*2*1 to get the numbers of ways to choose the committee.

Answer 9

so our answer is actually 8*7*6/(3*2*1) ?

Answer 10

yes, mathmate is correct, I left out that very important part! sorry. Since each combination of three people has six different permutations, you have to divide the entire thing by six.

Answer 11

okay thanks for the clarification!!