The expansion of $(a+b+c+d)^{20}$

Question

Let us consider the expansion of $$(a+b+c+d)^{20}.$$ Find:

• The coefficients of $a^{11}b^6c^2d$ and $a^{11}b^9$,
• The total number of terms of this expansion,
• The sum of all the coefficients.

Thank you for your help.

Answer 1

Observe that each term in the expansion of $$(a+b+c+d)^{20}=(a+b+c+d)(a+b+c+d) \cdots (a+b+c+d)$$ is obtained by taking one of the four terms $a,b, c$ or $d$ from each of the $20$ factors. Hence, the coefficient of $a^{11}b^6 c^2 d$ is the number of ways to choose $11$ of the $20$ factors for $a$ (which is ${20 \choose 11}$), times the number of ways to choose $6$ of the remaining $9$ factors for $b$ (which can be done in ${9 \choose 6}$ ways), times the number of ways to choose factors for $c$ (which can be done in ${3 \choose 2}$ ways), times the number of ways to choose the factors for $d$ (which can be done in ${1 \choose 1}$ ways). Hence, the coefficient in question is the product ${20 \choose 11}{9 \choose 6}{3 \choose 2}{1 \choose 1}$, which is equal to the multinomial coefficient $\frac{20!}{11! 6! 2! 1!}$. The coefficient of $a^{11} b^9$ can be obtained similarly.

To obtain the total number of terms in this expansion, observe that one term is of the form $\alpha a^7 b^7 c^6$ for some constant $\alpha$, and another is of the form $\beta a^6 b^6 c^6 d^2$. The former term corresponds to the partition $(7,7,6,0)$ of $20$ into $4$ parts, and the latter term corresponds to the partition $(6,6,6,2)$ into 4 parts. The number of ways to divide $20$ into $4$ parts is ${20+4-1 \choose 3}={23 \choose 3}$. To see why, place $20$ beads in a straight horizontal line, and you want to add three vertical lines (separator bars) to obtain the partition of $20$ into $4$ parts. The number of permutations of $20$ identical beads and $3$ identical bars is $\frac{23!}{3! 20!}$.

The sum of all coefficients in an expansion $\alpha a^7 b^7 c^6+ \beta a^6 b^6 c^6 d^2 \cdots$ is the sum $\alpha + \beta + \cdots$. Observe that the latter is obtained from the former by substituting $1$ for $a,b,c$ and $d$. Hence, we can substitute $1$ for $a,b,c$ and $d$ in the left hand side, ie in $(a+b+c+d)^{20}$, and we get $4^{20}$.

Answer 2

The first are just the multinomial coefficients $\binom{20}{11,6,2,1} = \frac{20!}{11! 6! 2! 1!}$ and $\binom{20}{11,9,0,0} = \frac{20!}{11!9!0!0!}$.

There is one term for every tuple $(a,b,c,d)$ with $a+b+c+d = 20$. where $a,b,c,d \ge 0$, all integers. This equals (e.g. see this wiki article) $\binom{23}{3}$.

To get the sum of all, just consider the multinomial theorem. Consider $(x_1 + x_2 + x_3 + x_4)^{20}$, which has all the same coefficients. To get the sum of them, fill in $x_1 = x_2 = x_3 = x_4 = 1$ to make all products of $x_i$ equal to 1, so we get $4^{20}$ as the answer.