Binomial Expansion Distinct Terms

Question

I have been stuck on a problem and I was wondering if someone could help me solve it. I am asked to find the number of distinct terms in the multivariate polynomial $f=(a+2b-c+3d)^{17}$ in which three of the powers of these terms are precisely even, and then determine the number of terms in which at least two of powers are greater than $6$. Does anyone know how to approach this problem? Is there is a formula that can be used to calculate these numbers?

Answer 1

We have to select $17$ times one element from $\{a,b,c,d\}$.

• Two of these elements have to occur at least $7$ times. We select exemplarily $a,b$ and consider \begin{align*} a^7(2b)^7(a+2b-c+3d)^3\tag{1} \end{align*}

• Since precisely three elements have to be even powers it follows from (1) that the only possible solutions from $(a+2b-c+3d)^3$ in this configuration are \begin{align*} a^8(2b)^8c\qquad \text{and}\qquad a^8(2b)^8(3d)\tag{2} \end{align*}

We observe from (1) we can select $\binom{4}{2}=6$ tuples from $\{a,b,c,d\}$ as we did with $a$ and $b$ . Then we have two ways to find valid choices according to (2). We conclude the number of distinct terms with the wanted properties is \begin{align*} \color{blue}{2\binom{4}{2}=12} \end{align*}