Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Yes, (1) is correct, assuming of course the student must solve exactly 8 of 10 problems, no more, no fewer. –  amWhy Apr 1 '13 at 15:11

2 Answers 2

up vote 0 down vote accepted

$(2)$ 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} $

share|improve this answer

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 $6$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.