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
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.
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
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)!] )
Misty - AEB is the same combination as ABE
Combinations pay no attention to order, permutations do.
Oh, sorry, I didn't catch my mistake. Ty
To determine the number of different three-person committees that can be formed from the five people (A, B, C, D, E), we can use the concept of combinations.
One way to set up a diagram to answer this question is by using a table or a grid. Here's an example of how you can set it up:
First, list all the possible choices for the first person in the committee (A, B, C, D, E) as column headers. Underneath each column header, write down the four remaining options for the second person, excluding the one chosen for the first person. Continue this pattern until you have accounted for all the remaining choices for the third person.
A B C D E
1 - B C D E
2 A - C D E
3 A B - D E
In this diagram, "-" represents that the same person cannot be chosen for that position, as each committee consists of three different people.
Now, count the number of possible committees by examining the entries in the diagram. In this case, there are 4 possible committees:
1. A, B, C
2. A, B, D
3. A, B, E
4. B, C, D
So, there are 4 different three-person committees that can be formed from these five people.