Permutations of 7 numbers less than 50

Question

How many solutions does the equation $a + b + c + d + e + f + g ≤ 50$ have if each variable must be a non-negative integer?

I feel that the answer is $50 \choose 7$?

but that seems far too simple.. Am I missing something huge?

Answer 1

The answer is $\binom{57}{7}$.

Take $57$ apples and pick $7$ of them out as borders. Then $50$ apples are left and the borders split them up in $8$ parts. $a$ stands for the number at the left side of the utmost left border, $b$ for the number between the utmost left border and its neighbour border, et cetera. Finally $g$ is the number between the most right border and its neighbour border. Together with the apples on the right side of the utmost right border they add up to $50$, so $a+\cdots+g\le50$.

Answer 2

I would look at using a generating function. Each letter $a, ..., g$ has the same constraint. So we model an individual constraint:

$$\sum_{i=0}^{\infty} x^{i} = \frac{1}{1-x}$$

Now we have seven elements, so we have our generating function $gen(x) = \frac{1}{(1-x)^{7}}$. And so using our binomial identities, we get the coefficient of $x^{50}$ as $\binom{50 + 7 - 1}{50}$.

Edit: I missed that you want the count for less than or equal to 50. So you can add up the binomial coefficients then: $$\sum_{i=0}^{50} \binom{i + 50 - 7}{i} $$