coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

find the coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$
(A)3400 (B)3410 (C)3420 (D)3430 (E)3440

so it would be $$x^{140} + ...... + 1$$
This requires binomial theorem and Multinomial theorem, but I'm not sure how to calculate it. Any tips or formula would be appreciate.
 A: $17$ can only be obtained by using two $5$s and one $7$ . These two $5$s can be obtained in $\binom{20}2$ ways which is $190$ and the $7$ can be got in from one of the remaining 18 brackets. So $190$ x $18$ = $3420$ is the answer.
A: Usage of the multinomial coefficient $(k_1, k_2, \cdots, k_n)$!:

$$
\big( 1 + x^5 + x^7\big)^{20} = \sum_{k_1=1}^{20} \sum_{k_2=1}^{20-k_1}
(k_1, k_2, 20 - k_1 - k_2)! x^{5k_1} x^{7k_2},
$$

where

$$
(k_1, k_2, \cdots, k_n)! = \frac{ (k_1 + k_2 + \cdots + k_n )! }
{ k_1! k_2! \cdots k_n!}.
$$

So we get $k_1=2$ and $k_2=1$, thus

$$
(2,1,17)! = \frac{(2+1+17)!}{
\begin{array}
{ccc}2! & 1! & 17!\\
\downarrow & \downarrow & \downarrow\\
2 \times 5 & 1 \times 7 & 17 \times 0
\end{array}
} = 3420.
$$

[change in order due to comment of @Henning Makholm]
A: $(1+x^5+x^7)^{20}=\{(1+x^5)+x^7\}^{20}$
$=(1+x^5)^{20}+\binom{20}1(1+x^5)^{20-1}(x^7)^1+\binom{20}2(1+x^5)^{20-2}(x^7)^2+\cdots+(x^7)^{20}$
So the required sum will be 
the coefficient of $x^{17}$ in $(1+x^5)^{20}$
$+\binom{20}1\cdot$ the coefficient of $x^{17-7}$ in $(1+x^5)^{20-1}$
$+\binom{20}2\cdot$ the coefficient of $x^{17-7\cdot2}$ in $(1+x^5)^{20-2}$
Clearly the exponent of $x$ in $(1+x^5)^n$ will be divisible by $5$
So, the first &  the last summand must be zero
Now for the coefficient of $x^{17-7\cdot1}$ in $(1+x^5)^{20-1},$
the $r+1$th term $\binom{19}r(x^5)^r=\binom{19}rx^{5r}$ and we need $5r=10\iff r=?$
A: So if you think about 
$$ (1 + x^5 + x^7)^{20} $$ 
That intuitively is just 
$$ ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) ... $$ 
Which can be expanded out term by term. By the Binomial Theorem as 
$$ (1 + x^5)^{20} (x^7)^0 + \begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 + \begin{pmatrix} 20 \\ 2\end{pmatrix}(1 + x^5)^{18}x^{14} ...$$
Now we can take each of the $$(1 + x^5)$$ terms and expand them as well. 
Note that the number 17 can be expressed as sum in terms of multiples of 5 and 7 as
$$ 10 + 7$$
If we consider say $15$ in the sum it's too big, and same for $14$ likewise arguments can be made to show that $5$ by itself won't add to a positive multiple of 7 to make 17. 
So that means every power of 17 is contained in the term
$$\begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 $$ 
Of our binomial expression. We need to find the coefficient of $x^{10}$ in 
$$ (1 + x^5)^{19} $$ 
Call that C. Then 
$$ C \begin{pmatrix} 20 \\ 1\end{pmatrix}$$ 
Is the answer. 
By Binomial Theorem:
$$  (1 + x^5)^{19}  = 1 + \begin{pmatrix} 19 \\ 1\end{pmatrix}x^5 + \begin{pmatrix} 19 \\ 2\end{pmatrix}x^{10} ... $$ 
So then $C = \begin{pmatrix} 19 \\ 2\end{pmatrix} $
So the answer then is 
$$\begin{pmatrix} 19 \\ 2\end{pmatrix} \begin{pmatrix} 20 \\ 1\end{pmatrix} $$
To get more advanced techniques (it is worthwhile to take a look at the multinomial theorems). They generalize these ideas for arbitrarily large sums raised to integer powers like the binomial theorem for 2 elements raised to a power.
Note that
$$\begin{pmatrix} 19 \\ 2\end{pmatrix} \begin{pmatrix} 20 \\ 1\end{pmatrix}  = \frac{19!}{2!17!} 20 = 19 \times 9 \times 20 = 3420$$
