There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
I see how you got that but my choices are
A. 6
B. 512
C. 56
D. 336
I don't understand it.
using the combination 8C3=8!/5!3!
8*7*6/3*2=56
option C
To find the number of different subcommittees possible, we can use the combination formula. The formula for combinations is given by:
nCr = n! / (r! * (n-r)!)
where:
n = total number of items
r = number of items to be chosen
In this case, we have 8 members on the board of directors and we want to form a subcommittee of 3 members. Therefore, we want to find the value of 8C3.
Using the combination formula:
8C3 = 8! / (3! * (8-3)!)
= 8! / (3! * 5!)
Now, let's calculate the factorial values:
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40,320
3! = 3 * 2 * 1 = 6
5! = 5 * 4 * 3 * 2 * 1 = 120
Substituting these values back into the combination formula:
8C3 = 40,320 / (6 * 120)
= 40,320 / 720
= 56
Therefore, there are 56 different subcommittees possible.
For 8 members, there are 8!/3! ways to choose a subcommittee of 3 members.
8!/3! = 8*7*6*5*4 = 6720