[Math] 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

combinatorics

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:

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

$= 931$

Best Answer

Method 2 should be $14\choose 4$-$8\choose 4$.

That will give you 931.

Related Solutions

Number of ways 24 students of 12 boys and 12 girls, how many ways can be separated into groups of 3 if all groups have to be mixed gender

You're idea is right, but it needs some refinement. First you have to select the $8$ boys in the first step and similarly the $8$ girls in the second one. Moreover for the last step we have $8!$ ways to do it. Indeed the first boy can be placed in $8$ groups, the next one in $7$ and so on.

Finally to avoid double counting you have to divide by $(2!)^8$. This is because each group has $2$ boys (or girls) and each of it can be used in the first step of creating the very same group.

Additionally you want to divide by $8!$, as I assume the group are undistinguishable. I believe that having placing boy $1,2$ and girl $1$ in the first group is same as placing them in the second one. Thus the final answer is:

$$\frac{\binom{12}{8} \cdot 8! \cdot \binom{12}{8} \cdot 8! \cdot 8!}{(2!)^8 \cdot 8!}$$

Maximum number of students such that they all have at least 3 out of 6 different answers.

You can model this as a maximum clique problem in a graph (equivalently, maximum independent set in the complement) with 64 nodes $\{0,\dots,2^6-1\}$ and an edge if the Hamming distance is at least 3. The maximum is 8:

EDIT:

Best Answer

Related Solutions

Number of ways 24 students of 12 boys and 12 girls, how many ways can be separated into groups of 3 if all groups have to be mixed gender

Maximum number of students such that they all have at least 3 out of 6 different answers.

Related Question