vongdn wrote:What is the number of different committees of 4 people that can be selected from a group of 10 people?

is that 10!/6!4! the answer?

diebeatsthegmat wrote:

vongdn wrote:What is the number of different committees of 4 people that can be selected from a group of 10 people?

is that 10!/6!4! the answer?

Yes, the answer is 210. But how did the factorials you selected relate to the information given in the problem? What was the rule for the solution?

Thanks for the info.

vongdn wrote:What is the number of different committees of 4 people that can be selected from a group of 10 people?

I use what some call the "slot method":

Draw a slot for each choice that is being made. Since we want 4 people on the committee, we draw 4 slots:

__ * __ * __ * __

We have 10 choices for the 1st slot (because we have 10 people to choose from).
We have 9 choices for the 2nd slot (because we used 1 person to fill slot 1, leaving us 10-1=9 choices for the 2nd slot).
We have 8 choices for the 3nd slot (because we used 2 people to fill slots 1 and 2, leaving us 10-2=8 choices for the 3rd slot).
We have 7 choices for the last slot (because we used 3 people to fill slots 1, 2 and 3, leaving us 10-3=7 choices for the last slot).

So far we have: 10 * 9 * 8 * 7

If we multiply the numbers above, we'll be counting the number of ways we can arrange 4 people from the 10 choices. For example, ABCD is a different arrangement from BCAD. The mathy word for this is permutation. There are 10*9*8*7 = 5040 ways to arrange 4 people from 10 choices.

But the problem above is asking how many groups can be made, and ABCD and BCAD are the same group (the mathy word for this is combination). To account for the duplicate combinations, we need to divide by (number of slots)!. This division will remove from our result all the duplicate combinations (so we don't overcount ABCD and BCAD as different combinations).

Since in this problem we have 4 slots, we divide by 4!:

10*9*8*7/4*3*2*1 = 210. So there are 210 ways to combine 4 people from 10 choices. Notice that the number of possible combinations is smaller than the number of possible permutations.

Hope this helps!