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Because at most, we can only have 1 ball in each box, we know that the partition we will be working with is $1+1+1+1+1$.

I know that if we take away the rule of at most 1 ball per box, the answer would be $8^5$ ways to distribute the balls to the boxes, but i'm not sure how to calculate this.

At first, my thought is to take $8^5$ and subtract every other scenario where a box has more than one ball, but I feel like that would take longer than just counting how many will have one each?

EDIT: The balls and boxes are both identical.

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    Are the boxes identical or distinguishable? – 2017-01-31
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    Are the balls distinguishable? – 2017-01-31
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    they are identical. I just added a comment, should have stated it earlier. – 2017-01-31
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    How is this problem different from choosing 5 boxes out of 8? – 2017-01-31
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    If both balls and boxes are identical, then there is only one distinct way. Put each indistinguishable ball into one each of the indistinguishable boxes. It doesn't matter which ball goes into which box; you can't tell the difference. – 2017-01-31
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    @GrahamKemp It says "at most" so you can empty boxes. – 2017-01-31
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    if boxes were distinguishable* then how would it work? – 2017-01-31
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    @Qtheplatypus But if the boxes are *indistinguishable*, how can you tell the arrangements apart? You just have 5 boxes with one ball and 3 boxes with no balls in no discernible order. – 2017-01-31
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    You can have an arrangement of "8 boxes with no balls" then "7 boxes with no balls and 1 box with a ball" etc. I can tell those arrangements apart. – 2017-01-31

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If both boxes are identical, then there is only one distinct way.   Put one ball into each of the indistinguishable boxes.   It does not matter which ball goes into which box when you can not tell the difference.

If the boxes are distinguishable, you need to select $5$ from $8$ boxes and put the balls into them.

If the balls are also distinguishable, also you need to count ways to rearrange the 5 balls in their selected positions.

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    thanks for the easy explanation! – 2017-01-31
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    graham can't we use stars and bars$ \sum_{i=1}^{8} x_i=8$ using the each $x_i\le1$ and defining each as $x_i+k_i'=1$, we get $\sum_{i=1}^{8} k_i'=3$ where k_i\ge1. and get $\comb{3+8-1}{8-1}$. where am i wrong if its not the answer then what does this calculate – 2017-01-31
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    the comb i meant $\binom{3+8-1}{8-1}$ – 2017-01-31
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    @Upstart There's an upper restraint on number of balls in any box (it is 1) that the basic "stars and bars" technique does not accomodate. – 2017-01-31
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    but i have stated that $x_i+k_i=1$ – 2017-01-31
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    @GrahamKemp did you also considered the cases where $n-boxes <=5$ can have $0$ balls because question stated an upper bound ? – 2017-11-24
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    @user292174 No the question stated that 5 identical balls *were to be* distributed into 8 identical boxes where *each box* can hold at most one ball. There is only one distinct way to do so. – 2017-11-24