How many different four-person committed can be formed from a group of six boys and four girls?
I don’t understand how to do this problem or set it up at all
This is a combination, not a permutation. That is, the order in which the people are chosen does not matter.
You have 6 boys and 4 girls, or you have 10 people.
Of those you have to choose groups of 4, and since no restriction is given,
C(10,4) or 10!/(6!4!) = 210
(hint: most scientific calculators have a key labelled nCr which will do the arithmetic for you)
Hey, cool, never thought to look for that key :)
To solve this problem, we can use the concept of combinations, which is a way to count the number of ways to select a certain number of items from a larger set. In this case, we want to count the number of ways to select four people from a group of six boys and four girls.
To set up the problem, we need to consider two scenarios: when all four selected people are boys and when at least one girl is included.
First, let's calculate the number of ways to select four boys from a group of six boys, since this scenario guarantees that there won't be any girls in the group. This can be calculated using combinations. The formula for combinations is:
nCr = n! / (r!(n-r)!)
where n represents the total number of items, and r represents the number of items to be selected.
For this scenario, we have n = 6 (number of boys) and r = 4 (number of boys to be selected). Substituting these values into the formula:
6C4 = 6! / (4!(6-4)!)
= 6! / (4!2!)
Simplifying:
6C4 = (6 * 5 * 4!)/(4! * 2 * 1)
= (6 * 5)/(2 * 1)
= 15
So there are 15 ways to select four boys.
Next, let's consider the scenario where at least one girl is included in the group. We can calculate this by subtracting the number of cases when no girls are included from the total number of possible combinations.
To calculate the total number of possible combinations, we need to consider selecting four people from a group of ten (six boys + four girls). This can be calculated using the combination formula as well:
10C4 = 10! / (4!(10-4)!)
= 10! / (4! * 6!)
Simplifying:
10C4 = (10 * 9 * 8 * 7 * 6!)/(4! * 6!)
Since 6! cancels out, we have:
10C4 = (10 * 9 * 8 * 7)/(4 * 3 * 2 * 1)
= 210
Now, we know that there are 15 ways to select only boys. So, to calculate the number of ways that include at least one girl, we subtract 15 from the total possible combinations:
Number of ways = 210 - 15
= 195
Therefore, there are 195 different four-person committees that can be formed from a group of six boys and four girls.