the class president plans to randomly select a committee of three people from three boys and five girls. How many committees are possible?
You are simply choosing 3 people from 8
that would be 8CHOOSE3 or C(8,3)
= 8!/(3!5!)
= 56
thanks
Since it is a combination and order does not matter. You would have 8 students taken 3 at a time.
MATH:
8C3= 8×7×6/3×2×1
=56 possible committees
By the way, the C in 8C3 does NOT stand for CHOOSE. It stands for combination.
To find out the number of possible committees, we need to use the concept of combinations.
In this case, the class president wants to select a committee of three people from a group of three boys and five girls. This means we need to select three people from the total eight candidates (3 boys + 5 girls).
The formula to calculate the number of combinations is given by the combination formula:
C(n, r) = n! / (r! * (n-r)!),
where n represents the total number of candidates and r represents the number of people to be selected for the committee.
In our case, n = 8 (boys + girls) and r = 3 (number of committee members).
Plugging the values into the formula:
C(8, 3) = 8! / (3! * (8-3)!)
Simplifying further:
C(8, 3) = 8! / (3! * 5!)
8! = 8 * 7 * 6 * 5!
5! = 5 * 4 * 3 * 2 * 1
Now we can substitute back into the formula:
C(8, 3) = (8 * 7 * 6 * 5!) / (3! * 5!)
The 5! terms in the numerator and denominator will cancel out:
C(8, 3) = (8 * 7 * 6) / (3! * 1)
3! = 3 * 2 * 1
Simplifying further:
C(8, 3) = (8 * 7 * 6) / (6)
Finally, we can calculate:
C(8, 3) = 8 * 7
C(8, 3) = 56
Therefore, the number of possible committees is 56.