finding the coefficient of ${t}^{20}$ in the expansion of ${({t}^{3}-3{t}^{2}+7t+1)}^{11}$

Question

I saw a question in my textbook.I tried to solve it using multinomial theorem.However,i stuck in somewhere.

The question is: find the coefficient of ${t}^{20}$ in the expansion of ${({t}^{3}-3{t}^{2}+7t+1)}^{11}$

My solution is: A general term of the expansion has the form $C(11;b_1,b_2,b_3,b_4)$ ${({t}^{3})}^{b_1}$${({-3t}^{2})}^{b_2}$${({7t})}^{b_3}$${(1)}^{b_4}$

In order to have a coefficient of ${t}^{20}$, we must have $b_1+b_2+b_3+b_4=11$ and $3b_1+2b_2+b_3=20$

I could not move foward in here ,because i could not reach the values of $b_1,b_2,b_3,b_4$. I obtained just equations like $b_3=20-2b_2-3b_1$ and $b_4=2b_1+b_2-9$.

Can you help me for the rest ? Thank you for your helps..

Answer 1

Enumerating the tuples is not very hard. As we see $b_j$s are non negative and $b_1\leq6$, so we pick successive legal values of $b_1$, which will limit the upper value of $b_2$: for example if we pick $b_1=3$, we get $2b_2+b_3=11$, which limits the upper value value of $b_2$ at $5$. We then pick successive values of $b_2$ and so on. $b_4$ is just the residue needed to sum up to $11$. We truncate wherever the sum of $b_1, b_2, b_3$ exceeds $11$.
So here's a table where I list out the tuples in this order.

$$\left[\begin{array}{ccc|c|c|c} b_1 & b_2 & b_3 & b_4 & \frac{11!}{b_1!b_2!b_3!b_4!} & (-3)^{b_2}7^{b_3}\\ 6 & 1 & 0 & 4 & 2310 & -3\\ 6 & 0 & 2 & 3 & 4620 & 49\\ \hline 5 & 2 & 1 & 3 & 27720 & 63\\ 5 & 1 & 3 & 2 & 27720 & -1029\\ 5 & 0 & 5 & 1 & 2772 & 16807\\ \hline 4 & 4 & 0 & 3 & 11550 & 81\\ 4 & 3 & 2 & 2 & 69300 & -1323\\ 4 & 2 & 4 & 1 & 34650 & 21609\\ 4 & 1 & 6 & 0 & 2310 & -352947\\ \hline 3 & 5 & 1 & 2 & 27720 & -1701\\ 3 & 4 & 3 & 1 & 46200 & 27783\\ 3 & 3 & 5 & 0 & 9240 & -453789\\ \hline 2 & 7 & 0 & 2 & 1980 & -2187\\ 2 & 6 & 2 & 1 & 13860 & 35721\\ 2 & 5 & 4 & 0 & 6930 & -583443\\ \hline 1 & 8 & 1 & 1 & 990 & 45927\\ 1 & 7 & 3 & 0 & 1320 & -750141\\ \hline 0 & 10 & 0 & 1 & 11 & 59049\\ 0 & 9 & 2 & 0 & 55 & -964467\\ \end{array}\right]$$

The grunt part is of course computing the multinomial coefficient and the product of powers. But once you do a sum of their pairwise products you get:

Coefficient of $t^{20}$ = $-7643472342$

I verified the answer programmatically and at this polynomial expansion calculator

Answer 2

From $3b_1+2b_2+b_3=20$ follows equal parity of $b_1$ and $b_3.$
Then $b_1+b_2+b_3+b_4=11$ says that $b_2$ and $b_4$ have opposite parity.
Shortly: three of the coefficients are odd and one is even, or vice versa.

Now, I suggest to create a table and find systematically all quadruples, e.g. $(0,10,0,1),(0,9,2,0)...$ They are not so many.