Posted by **chris** on Monday, January 21, 2008 at 5:41pm.

A three person committee is formed with the following people to be considered: A, B, C, D, E. How many different three-person committees can be formed from these five?

what is the best way to set up a diagram to answer this question? can you show how to do it or tell what the answer is? thanks

- math -
**Misty**, Monday, January 21, 2008 at 5:54pm
ABC

ABD

ABE

ACD

ACE

ADE

AEB

AEC

AED

To me it would be the first letter and add 2 letters to it, not repeating the first. Like shown above. Then go on to the next letter and do the same. This will take some time but should get you your answer.

- math -
**Damon**, Monday, January 21, 2008 at 6:05pm
I can tell what the answer is :)

The number of combinations of r people from a group of n people is

C = n! / [r!(n-r)!]

C = 5!/[3!*2!] = 5*4/2 = 10

Now what you asked:

the first person A has a choice of 6 groups to join

it could be

ABC

ABD

ABE

ACD

ACE

ADE

The second person B has choice of 3 new groups to join because we have already done the ones with A

BCD

BCE

BDE

The third person C can only join with D and E

CDE

so we have

6+4+1 which is Lo and Behold =10

- math -
**Damon**, Monday, January 21, 2008 at 6:37pm
By the way, that answer, 10, is the same as the number of combinations of two out of five people

5!/(3!*2!) = 5!/(2!*3!) of course.

When my son was taking algebra 2 I wandered into the high school for parents' night.

The teacher asked five of us parents up to the front of the room and asked everyone how many times we had to shake hands for all five to shake hands with everyone else of the five.

Luckily there was a blackboard behind me :) I did it basically with the list way because I did not think it was the moment for factorials or Pascal's triangle (which you should look up for another way to get that n!/[r!(n-r)!] )

- math -
**Damon**, Monday, January 21, 2008 at 6:52pm
Misty - AEB is the same combination as ABE

Combinations pay no attention to order, permutations do.

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