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.
 A: 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 $6$.
A: $(2)$
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} $
