# Combinatorics problem

A student has to solve $8$ of $10$ problems in his test.

How many options/possibilities does he have

1)all in all?

2)if he has to solve one of the 2 first problems?

3)if he have to solve at least 4 of the 6 first problems?

What I have:

1)$\dbinom{10}{8}=45$

2)

3)

I don't really know how to proceed for question 2) & 3) and if 1) is correct.

• Yes, (1) is correct, assuming of course the student must solve exactly 8 of 10 problems, no more, no fewer. – Namaste Apr 1 '13 at 15:11

$(2)$ If he chooses the 1st Question, he can choose the rest $7$ from the rest $9$ Question in $\binom 97$ ways

If he chooses the 2nd Question, he can choose the rest $7$ from the rest $9$ Question in $\binom 97$ ways

If he chooses the 1st and the 2nd Questions, he can choose the rest $6$ from the rest $8$ Question in $\binom 86$ ways (This is the intersection of the first two cases)

So, the number of combinations will be $2\cdot\binom 97-\binom 86=2\cdot\binom 92-\binom 82=72-28=44$ as $\binom nr=\binom n{n-r}$

$(3)$

If he chooses exactly $4$ out of $6,$( which he can do in $\binom 64$ ways), he can choose the rest $4$ from the rest $4$ in $\binom44$way.

So, this gives us $\binom 64\cdot\binom{10-6}{8-4}$ combinations

And so on

So, the required combination will be $\binom 64\cdot\binom{10-6}{8-4}+\binom 65\cdot\binom{10-6}{8-5}+\binom 66\cdot\binom{10-6}{8-6}$

Assuming that the student is lazy and doesn't consider the options of solving $9$ or $10$ problems, your answer to $1)$ is correct.

For $2)$, either exactly $1$ of the first two problems can be solved, which yields $\displaystyle\binom21\binom87=16$ options, or both of the first two problems can be solved, which yields $\displaystyle\binom22\binom86=28$ options, for a total of $16+28=44$ options.

The answer to $3)$ is the same as for $1)$, since it's impossible to solve $8$ of $10$ problems without solving $4$ of the first $6$.