# How many selection ways are possible?

There are 6 girls and 5 boys. You need to make a team of 4 persons. In how many ways can you form this team?

One way is to select 0,1,2,3,4 girls in team, but this is a very lengthy process. Is there any fast solution?

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In problems like this, it is often very helpful to give two or three specific examples of the different things being counted. This clarifies what you're considering "different" and what you're considering "not different". – vadim123 Jun 4 '13 at 14:13
How is this different from the question "There are 11 kids and you need to make a team of 4 people"? Are there any requirements on the team having some members of each gender, etc.? – Peter Košinár Jun 4 '13 at 14:13
Please do not change questions after answers are posted. It makes answers non-responsive. I already answered the new one because it seemed the problem was because there were only two categories. You might look at stars and bars for a more general case. – Ross Millikan Jun 4 '13 at 16:42
ok. got it. but i dnt get solution. – Nitin Bansal Jun 4 '13 at 16:47

If the girls and boys are distinguishable, you really have eleven players to choose four from. Writing using the binomial coefficient this is just ${11 \choose 4}$. If they are not, you are just asking how many ways there are to split 4 into two ordered integers, and you have listed the five ways.
Added: Yes, as I explained, you can just find the maximum and minimum, the the count is $\max-\min+1$ where the $+1$ comes because you have both ends. You don't have to think about the numbers in the middle. For a more complicated case, suppose you have lots of red, blue, and yellow jelly beans and want the number of ways to select $10$ of them. You can say that the number of red ones can range from $0$ to $10$. Given $r$ reds, there are $11-r$ ways to choose the yellows and blues, so the total is $\sum_{r=0}^{10} (11-r)=11\cdot 11-\frac 12\cdot 11\cdot 12=121-66=55$
You can just say the minimum number of girls is $0,$ the maximum is $4,$ so the range is $4-0+1=5$. Is that what you are looking for? – Ross Millikan Jun 4 '13 at 14:18