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If n is less than 10m it is impossible as not all of them can have at least 10 chocolates.

So as long as there are n ≥ 10m chocolates, 10 chocolates are given to each mathematician and the remaining extra x = n - 10m chocolates need to be distributed among the m mathematicians which can be done in:

m - 1 + xCm - 1 = (m - 1 + x)!/(m - 1)!x!

ways.

Using only m and n, this becomes:

n - 9m - 1Cm - 1 = (n - 9m - 1)!/(m - 1)!(n - 10m)!

ways.

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14y ago

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