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This is my attempt at the problem:

There are $\binom{10+62-1}{62}$ ways to put the 62 white balls into the 10 distinguishable boxes. I am not sure how count the number of ways to put 8 distinguishable numbered balls into 10 distinguishable boxes. If I did though, I would multiple it by $\binom{10+62-1}{62}$ and that would be my final answer.

The answer in the book is $\binom{10+62-1}{62} 10^8$. Where did the $10^8$ come from?

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    Each distinguishable ball has $10$ choices, and they can be made independently . – 2017-02-10
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    @lulu Wow, such a slight modification to the way I was thinking about it but it makes total sense now. Thanks! – 2017-02-10

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