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Find number of solutions
$$ x_1+x_2+x_3+x_4+x_5+x_6+x_7 = 7 text such that forall_i x_i in left0,1,2right$$
I know how I can do this when I don't have restriction $forall_i x_i in left0,1,2right$:
$$ ooooooooooooo text n+(k-1) = 7 + (7-1) = 13 balls $$
$$ oo||o|oo|o|o| text k-1 = 6 balls I replace with sticks $$
and I have $$ 2 + 0 + 1 + 2 + 1 + 1 + 0 = 7 $$
I can do this in $$ binomn+k-1k = binom137 $$ ways. But how to deal with additional restriction?

The number of unique combinations of numbers summed to achieve $7$ in such a way are
$$1,1,1,1,1,1,1$$
$$2,1,1,1,1,1$$
$$2,2,1,1,1$$
$$2,2,2,1$$
So the total number of solutions is given by
$$frac7!7!+frac7!5!+frac7!3!cdot2!cdot2!+frac7!3!cdot3!=393$$

Hint: The answer is the coefficient of $x^7$ in $(1 + x + x^2)^7$.

Hint: You want the coefficient of $x^7$ in
$$
(1+x+x^2)^7=left(frac1-x^31-xright)^7=(1-x^3)^7times (1-x)^-7
$$
Now, $(1-x^3)^7$ and $ (1-x)^-1$ are the generating functions of two nice series, $a_n$ and $b_n$; can you find them? Once you do, since you want the convolution of these two series:
$$
sum_k=0^7 a_kb_n-k.
$$
Furthermore, you will find that $a_k$ equals zero unless $k$ is a multiple of $3$, so that the above summation is has only three nonzero terms and is therefore easily computable by hand.

The "closed form answer" for the number of ways to assign $x_1, x_2, cdots ,x_k$ such that $forall n : x_n in 0,1,2$ and $sum_n=1^k x_n = k$ is, for odd $k$
$$
F^-frack2, -frack-12_1(4)
$$
and for even $k$
$$
F^-frack-12, -frack2 _1(4)
$$
These $F^a,b_c$ are hypergeometric functions.

This is obtained by letting $n$ be the number of $2$s used and doing
$$
sum_n=0^leftlfloorfrack2rightrfloor binomknbinomk-nk-2n
$$
and using the techniques put forth in Concrete Mathematics.

From the theory of Generating Functions it's clear the answer boils down to finding the coefficient of $x^7$ in $(1 + x + x^2)^7$,

Write out the equivalent of Pascal's Triangle for the Trinomial Coefficients, or look it up, or write a quick program to generate them (each term is the sum of the three terms, above left, directly above, above right)
$$1$$
$$1 : 1 : 1$$
$$1: 2: 3: 2: 1$$
$$1: 3: 6: 7: 6: 3: 1$$
$$1: 4: 10: 16: 19: 16: 10: 4: 1$$
$$1: 5: 15: 30: 45: 51: 45: 30: 15: 5: 1$$
$$1: 6: 21: 50: 90: 126: 141: 126: 90: 50: 21: 6: 1$$
$$1: 7: 28: 77: 161: 266: 357: 393: 357: 266: 161: 77: 28: 7: 1 $$
The number sought is the central coefficient in Row 7, the 393.