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

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 $4^{20}$.  
