Clarification about Combination problem 50 people consisting of 25 boys and 25 girls are sending a group of 20 people to camp. How many was can 20 people be constructed so that there are is at least 1 girl in the group?
I'm still trying to understand combinations. I believe the answer would just be $25 \choose 1$. Is this correct? The only thing I'm not sure about is how the 19 or less boys would come into play here.
 A: Let us count the number of groups without any girls: there are $25 \choose 20$ possibilities (just pick 20 boys out of 25).
Let us count the total number of possible groups: here are $50 \choose 20$ possibilities (just pick 20 kids out of 50).
Consequently you have ${50 \choose 20} - {25 \choose 20}$ groups with at least one girl.
Generally when you have to count the number of elements in a set such that at least 1 element verifies a property it is easier to count the number of elements such that 0 element verifies the property and then compute the difference with the total number of elements in your set.
A: You need "at least 1 girl". From a probability standpoint, P(at least one girl) = 1 - P(no girls)
For combinatorics, that would mean there can be 19 boys and 1 girl, or 18 boys and 2 girls, or ... up to 0 boys and 20 girls.
${25 \choose 19}{25 \choose 1}$ is the number of ways to choose 19 boys from 25 and 1 girl from 25. The total number of ways to choose at least 1 girl would then be:
$${25 \choose 19}{25 \choose 1} + {25 \choose 18}{25 \choose 2} + \ldots + {25 \choose 0}{25 \choose 20}$$ 
