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I can across this question:

Dillian has 5 pieces of paper, each with a different math problem. In how many ways can he give these problems to his 10 friends where each friend can receive more than one problem?

I am a little confused because I do not think the answer is simply $10^5$. What am I doing wrong?

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    $\begingroup$ It looks like $10^5$ to me: for each paper he has $10$ choice of friend to whom to give it, and there are no restrictions. $\endgroup$ – Brian M. Scott May 21 at 16:07
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How many ways can he give the 1st piece of paper to one of his friends? 10.

How many ways can he give the 2nd piece of paper to one of his friends? 10.

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How many ways can he give the 5th piece of paper to one of his friends? 10.

Each of these decisions is independent of each of the other.

So there are $10 \times 10 \times \dotsb \times 10$ ways of distributing his problems.

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He can give each problem to his friends in 10 different ways and since there are 5 problems the answer is $10^5$.

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  • $\begingroup$ Yes that is right.Apologies $\endgroup$ – Monocerotis May 21 at 16:35

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