Fun "facts" about integers

Probability Level 4

For non-negative integers $a,b,c$ with their sum of 10, let the triplets $(a_1,b_1,c_1), (a_2,b_2, c_2) , \ldots, (a_n, b_n, c_n)$ denote the ordered triplets of $(a,b,c)$ such that $\frac{10!}{a!b!c!}$ is an integer.

Find the value of $\displaystyle \sum_{j=1}^n \frac{10!}{(a_j)!(b_j)!(c_j)!}$ .

The answer is 59049.

1 solution

Abdelhamid Saadi
May 26, 2015

From the Mutlinomial formula

for $n = 1$ and $x_1 = x_2 = x_3 = 1$

The result is equal to $(1 + 1 + 1)^{10} = 59049$ .

QED

I believe you mean ${\left(1+1+1\right)}^{10}$ , right?

Dylan Pentland - 6 years ago

You are right.

A typing error . Fixed.

Thank you, Dylan

Abdelhamid Saadi - 6 years ago

Unless I'm mistaken, surely the sum should be a multiple of 10? a,b,c are defined to have a sum of 10, so all possible sets have an individual sum of 10, therefore the value of the expression given is 10*number of ordered triples, which I got to be 36 (without 0s) or 66 (allowing 0s), making the answer 360 or 660 depending on whether 0 is counted as a non-negative integer or not. Please correct me if i'm wrong.

Stewart Feasby - 6 years ago

The content of the problem has changed.

It was the sum of $\frac{10!}{a!b!c!}$ for $a + b + c = 10$

The solution for the new version is 660.

Abdelhamid Saadi - 6 years ago

Fun "facts" about integers

The answer is 59049.

1 solution

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