# There are 14 students: 8 girls and 6 boys. In how many ways can you make a 4-student committee which has at least one boy?

In a group of $14$ students there are $8$ girls and $6$ boys. Determine the number of ways that a committee of $4$ students which has at least $1$ boy can be chosen from the group.

Here is what I have so far:

Method $1$

If there is $1$ boy: $(^6C_1)\cdot (^8C_3) = 336$

If there are $2$ boys: $(^6C_2)\cdot (^8C_2)= 420$

If there are $3$ boys: $(^6C_3)\cdot (^8C_1) = 160$

If there are $4$ boys: $(^6C_4)\cdot (^8C_0)= 15$

$336+420+160+15 = 931$

I was confident in this answer, until I tried solving another way.

Method $2$

number of comittees with at least one boy = total number of committees possible – number of comittees with zero boys

$=(^{14}C_4)\cdot (^8C_8) = 1000$

Why are my two answers different?

## EDIT:

Method $2$ should be as follows:

$= (^{14}C_4)-(^8C_4)$

$= 931$

• Shouldn't method 2 be (14C4)-(8C4)? That might be 1000 and you just made a typo. I don't have a calculator handy. – turkeyhundt Dec 13 '14 at 4:46
• Ah, yes! No, it was my mistake. Thanks for your help. – McB Dec 13 '14 at 4:48
• @turkeyhundt is right, $\binom{14}{4}-\binom{8}{4}=931$. – Jack D'Aurizio Dec 13 '14 at 4:48

Method 2 should be $14\choose 4$-$8\choose 4$.
I think in the last step, you wanted to write $^{14}C_4-^8C_8=1000?$ In that case, why $^8C_8?$ From $8$ girls, you are choosing $4$ girls, so you should subtract $^8C_4$. Then you get 931.