Can someone explain how to do this:
There are 100 boys , 80 girls, and 20 teachers.
Out of these, how many ways can 2 girls, 2 boys and 1 teacher (all together in a group of 5) be chosen?
How many ways can 2 girls be chosen from 80?
n = 80
r = 2
C(n,r) = n!/[r!(n-r)!]
= 80!/[2!(78!)] 80*79/2 = 3160
then 2 boys from 100
100*99/2 = 4950
then 1 teacher from 20 is obviously 20
I now have 3160 groups of girls, 4950 groups of boys and 20 groups of teachers.
The ways of getting a particular group
of girls are 3160
the ways of combining that group of girls with a particular group of boys are 3160*4950
then the ways of combining a particular group of girls and boys with a particular teacher are 3160*4950*20
To determine the number of ways to choose 2 girls, 2 boys, and 1 teacher from the given group, we can use the concept of combinations.
First, let's calculate the number of ways to choose 2 girls from 80 and 2 boys from 100. We'll use the formula for combinations, denoted as "C(n, r)" where n is the total number of items and r is the number of items we want to choose.
The number of ways to choose 2 girls from 80 can be calculated as C(80, 2) = 80! / (2! * (80-2)!) = 80! / (2! * 78!) = (80 * 79) / (2 * 1) = 3,160.
Similarly, the number of ways to choose 2 boys from 100 can be calculated as C(100, 2) = 100! / (2! * (100-2)!) = 100! / (2! * 98!) = (100 * 99) / (2 * 1) = 4,950.
Now, let's calculate the number of ways to choose 1 teacher from 20, which is simply 20.
Multiplying these results together, we find the total number of ways to choose 2 girls, 2 boys, and 1 teacher is 3,160 * 4,950 * 20 = 313,600,000.
Therefore, there are 313,600,000 ways to choose 2 girls, 2 boys, and 1 teacher from the given group.