A company's board of directors wants to form a committee of
3
of its members. There are
4
members to choose from. How many different committees of
3
members could possibly be formed?
Please try the method that bobpursley showed you.
https://www.jiskha.com/display.cgi?id=1502046439
What is C(4,3) or 4!/(3!1!) ??
To find the number of different committees that can be formed, we can use the combination formula, also known as "n choose k".
In this case, we have 4 members to choose from, and we want to form a committee of 3 members. Therefore, we can calculate the number of different committees as follows:
4 choose 3 = 4! / (3! * (4-3)!) = 4! / (3! * 1!) = (4 * 3 * 2) / (3 * 2 * 1) = 4
So, there are 4 different committees of 3 members that could possibly be formed.
To calculate the number of different committees that can be formed from a group of members, we can use the concept of combinations. The formula for combinations is:
C(n, r) = n! / (r! * (n-r)!)
Where:
- n is the total number of members in the group
- r is the number of members needed for the committee
- ! represents the factorial operation, which means multiplying a number by all the positive integers less than it.
In this case, n = 4 (total number of members) and r = 3 (number of members needed for the committee). Plugging these values into the formula, we can calculate the number of different committees:
C(4, 3) = 4! / (3! * (4-3)!)
= 4! / (3! * 1!)
= (4 * 3 * 2 * 1) / (3 * 2 * 1 * 1)
= 24 / 6
= 4
Therefore, there are 4 different committees of 3 members that can be formed from a group of 4 members.