Q.1.26 - A First Course in Probability

Question

Expand${\left(x_1+2x_2+3x_3\right)}^4$.

Step-by-Step Solution

Answer

${\left(x_1+2x_2+3x_3\right)}^4=81{x_3}^4+216x_2{x_3}^3+216{x_2}^2{x_3}^2+96{x_2}^3{x}_3+16{x_2}^4+108x_1{x_3}^3+216x_1{x}_2{x_3}^3+144x_1{x_2}^2{x}_3+36x_1{x_2}^3+54{x_1}^2{x_3}^2+72{x_1}^2{x}_2{x}_3+24{x_1}^2{x_2}^2+12{x_1}^3{x}_3+8{x_1}^3{x}_2+{x_1}^4$

Step 1 Given Information.

${\left(x_1+2x_2+3x_3\right)}^4$

Step 2 Explanation.

Recognize the multinomial theorem:

$$\begin{gathered} {\left(x_1+x_2+\dots x_k\right)}^n=\sum _{\backslash \mathrm{substack}\left(n_1,n_2,\dots ,n_k\right): \\ {n}_1+n_2,\dots ,n{n}_k=n \\ {n}_1,n_2,\dots ,n_k\in {\mathrm{\mathbb{N}}}_0}\left(n_1,n_2,\dots ,n_k\right)x_1^{n_1}{x}_2^{n_2}\dots x_k^{n_k} \end{gathered}$$

So the $n$-th power of a sum will be a sum of products of $x_i^{n_i}$with nonnegative vector of integers$\left(n_1,n_2,\dots ,n_k\right)$. Here, every product associated with the vector $\left(n_1,n_2,\dots ,n_k\right)$means that we chose$x_i$ from$n_i$ factors in the initial product$\left(x_1+x_2+\dots x_k\right)$.

$\left(x_1+x_2+\dots x_k\right)\cdots \dots \left(x_1+x_2+\dots x_k\right)$and that can be done in$\left(\begin{array}{c}n\\ n_1,n_2,\dots ,n_k\end{array}\right)$ ways. Dividing

$n$factors into $n_1$from which to take $x_1,n_2$from which take $x_{2,}$and so on.

Step 3 Explanation.

$$\begin{gathered} {\left(x_1+2x_2+3x_3\right)}^4=\sum _{\backslash \mathrm{substack}\left(n_1,n_2,n_3\right): \\ {n}_1+n_2+n_3=4 \\ {n}_1,n_2,n_3\in {\mathrm{\mathbb{N}}}_0}\left(\begin{array}{c}4\\ n_1,n_2,n_3\end{array}\right)x_1^{n_1}{x}_2^{n_2}{x}_3^{n_3} \end{gathered}$$

Nonnegative integer vectors$\left(n_1,n_2,n_3\right)$ that sum up to $4$are$\left(\begin{array}{ccc}0,& 0,& 4\end{array}\right)$.

$$\begin{gathered} (0,1,3),(0,2,2),(0,3,1),(0,4,0),(1,0,3),(1,1,2),(1,2,1) \\ (1,3,0),(2,0,2),(2,1,1),(2,2,0),(3,0,1),(3,1,0),(4,0,0) \end{gathered}$$

This agrees with the result of the chapter$1.6$ which states that there are $\left(\begin{array}{l}6\\ 2\end{array}\right)=15$different nonnegative integer triplets that sum up$4$.

Separate calculation of the multinomials:

$\left(\begin{array}{ccc}& 4& \\ 4,& 0,& 0\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 0,& 4,& 0\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 0,& 0,& 4\end{array}\right)=\frac{4!}{0!0!4!}=1$

$\left(\begin{array}{ccc}& 4& \\ 0,& 1,& 3\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 0,& 3,& 1\end{array}\right)=...=\left(\begin{array}{ccc}& 4& \\ 3,& 1,& 0\end{array}\right)=\frac{4!}{0!1!3!}=4$

$\left(\begin{array}{ccc}& 4& \\ 0,& 2,& 2\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 2,& 2,& 0\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 2,& 0,& 2\end{array}\right)=\frac{4!}{0!2!2!}=6$

$\left(\begin{array}{ccc}& 4& \\ 1,& 1,& 2\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 1,& 2,& 1\end{array}\right)=\left(\begin{array}{ccc}& 4& \\ 1,& 1,& 2\end{array}\right)=\frac{4!}{1!1!2!}=12$.