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There are 3 sections in a question paper each containing 5 questions. A candidate has to solve only 5 questions, choosing at least one question from each section. In how many ways can he make his choice?

I have thought of a solution but I am over counting the number of ways.

No. of ways to choose one question from each section = (5C1)^3 No. of questions remaining = 12 No. of ways to pick 2 questions from remaining 12 questions = 12 * 11 Total number of ways = (5C1)^3 * 12 * 11

Can somebody tell me where I'm going wrong

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  • $\begingroup$ What is the answer? $\endgroup$ – callculus Oct 19 '18 at 16:18
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He can choose the following combinations of questions from each section:

$(2,2,1);(2,1,2);(1,2,2);(3,1,1);(1,3,1);(1,1,3)$

Each of the first three combinations has $\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}$ ways. And each of the next three combinations has $\binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}$ ways.

In total he can make $3\cdot \left(\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}+ \binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}\right)=2250$ choices.


For every section we have one path. We choose 3 question from each section. For every path we have to add ($\text{not multiply}$) the ways

$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)$

Now we make a case decision.

a) We choose 2 questions from one (other) section and 2 questions from the remaining section.

$ \binom{5}{2}\cdot \binom{5}{2}$

b) We choose 1 questions from one (other) section and 3 questions from the remaining section.

$ \binom{5}{1}\cdot \binom{5}{3}$

Therefore in total we have

$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)\cdot \left(\binom{5}{2}\cdot \binom{5}{2}+ \binom{5}{1}\cdot \binom{5}{3}\right)=2250$

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  • $\begingroup$ can you tell me where I'm going wrong? $\endgroup$ – Rishabh Khandelwal Oct 19 '18 at 16:22
  • $\begingroup$ One problem is $ \binom{5}{1}^3$. It must be $ \binom{5}{1}^1+\binom{5}{1}^1+ \binom{5}{1}^1$ $\endgroup$ – callculus Oct 19 '18 at 16:44
  • $\begingroup$ Still answer won't match $\endgroup$ – Rishabh Khandelwal Oct 19 '18 at 16:55
  • $\begingroup$ What is the answer? $\endgroup$ – callculus Oct 19 '18 at 16:58
  • $\begingroup$ 2250 is the correct answer $\endgroup$ – Rishabh Khandelwal Oct 19 '18 at 17:00

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