A group of friends consist of 3 girls and 5 boys. If that group of friends is separated into a group containing 1 girl and 2 boys, how many combinations of the original 8 friends could be made into the smaller group?
There three ways of choosing 1 girl out of three (3C1) and (5C2) ways of choosing 2 boys out of 5.
Total number of ways
= (3C1)(5C2)
= (3)(10)
= 30
To determine the number of combinations of the original group that can be made into the smaller group, we need to utilize the concept of combinations and apply it to the given scenario.
We know that there are 3 girls and 5 boys in the original group of friends. We need to select 1 girl and 2 boys from this group.
The number of ways to select 1 girl from a group of 3 can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
In this case, n = 3 (number of girls) and r = 1 (number of girls to be selected), so we have:
C(3, 1) = 3! / (1!(3-1)!) = 3! / (1! * 2!) = 3 / (1 * 2) = 3 / 2 = 3
Therefore, there are 3 ways to select 1 girl from the group.
Next, we need to select 2 boys from a group of 5. Using the combination formula again:
C(n, r) = n! / (r!(n-r)!)
Here, n = 5 (number of boys) and r = 2 (number of boys to be selected), so we have:
C(5, 2) = 5! / (2!(5-2)!) = 5! / (2! * 3!) = (5 * 4 * 3!) / (2! * 3!) = (5 * 4) / 2 = 20 / 2 = 10
Therefore, there are 10 ways to select 2 boys from the group.
To determine the total combinations, we multiply the number of ways to choose a girl (3) by the number of ways to choose 2 boys (10):
Total combinations = 3 * 10 = 30
Therefore, there are 30 combinations of the original 8 friends that can be made into the smaller group containing 1 girl and 2 boys.