Asked by Twg
There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
Answers
Answered by
MathMate
For 8 members, there are 8!/3! ways to choose a subcommittee of 3 members.
8!/3! = 8*7*6*5*4 = 6720
8!/3! = 8*7*6*5*4 = 6720
Answered by
Twg
I see how you got that but my choices are
A. 6
B. 512
C. 56
D. 336
I don't understand it.
A. 6
B. 512
C. 56
D. 336
I don't understand it.
Answered by
guash
using the combination 8C3=8!/5!3!
8*7*6/3*2=56
option C
8*7*6/3*2=56
option C
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