The choice A,B,C and D,E,F, should not be considered different than D,E,F and A,B,C.
2007-10-01 06:17:58 · 4 answers · asked by Delores S 1 in Education & Reference ➔ Higher Education (University +)
First -- think about the way you can choose one group of three people from a group of six. The number of combinations is found using the well known formula for choosing K things from N -- the binomial coefficient:
Combin(k,n) = n!/[k!*(k-n)!
In this case -- it is:
6!/[(3!*(6-3)!) = 20.
But 20 is not our answer. There are 20 ways to choose one team -- but if we are choosing two teams, we have to realize that we are duplicating every possibility twice (picking ABC is the same as picking DEF). Therefore, the answer is half of 20 -- which is ten.
2007-10-01 07:22:43 · answer #1 · answered by Ranto 7 · 1⤊ 0⤋
6
2007-10-01 06:25:31 · answer #2 · answered by Anonymous · 0⤊ 0⤋
abc def
abd cef
abe cdf
abf cde
acd bef
ace bdf
acf bde
ade bcf
adf bce
aef bcd
10 is correct!
2007-10-01 06:37:23 · answer #3 · answered by Anonymous · 0⤊ 0⤋
10
ABC DEF
ABD CEF
ABE CFD
ABF CED
ACD BEF
ACE BDF
ACF BED
BCD AEF
BCE ADF
BCF AED
2007-10-01 06:31:20 · answer #4 · answered by Eric D 3 · 1⤊ 0⤋