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