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Expert-verified Found in: Page 421 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Find the expansion of ${\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}^{\mathbf{4}}$${\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}^{\mathbf{4}}$a) using combinatorial reasoning, as in Example b) using the binomial theorem.

${x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}$

See the step by step solution

## Step 1: Formula for Binomial theorem

Binomial theorem is: ${\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}^{\mathbf{n}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathbf{ }}\left(\begin{array}{l}n\\ j\end{array}\right){{\mathbit{x}}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{{\mathbit{y}}}^{{\mathbf{j}}}$

## Step 2: Use Distributive property and Binomial theorem

Given: $\left(x+y{\right)}^{4}$

(a)

Write the given statement as a product and then use the distributive property

$\begin{array}{r}\left(x+y{\right)}^{4}=\left(x+y\right)\left(x+y\right)\left(x+y\right)\left(x+y\right)\\ \left(x+y{\right)}^{4}=\left({x}^{2}+xy+xy+{y}^{2}\right)\left({x}^{2}+xy+xy+{y}^{2}\right)\end{array}$

$\begin{array}{r}\left(x+y{\right)}^{4}=\left({x}^{2}+2xy+{y}^{2}\right)\left({x}^{2}+2xy+{y}^{2}\right)\\ \left(x+y{\right)}^{4}={x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{4}\\ \left(x+y{\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}$

(b)

Use the binomial theorem:

$\left(x+y{\right)}^{4}=\left(\begin{array}{l}4\\ 0\end{array}\right){x}^{4}{y}^{0}+\left(\begin{array}{l}4\\ 1\end{array}\right){x}^{3}{y}^{1}+\left(\begin{array}{l}4\\ 2\end{array}\right){x}^{2}{y}^{2}+\left(\begin{array}{l}4\\ 3\end{array}\right){x}^{1}{y}^{3}+\left(\begin{array}{l}4\\ 4\end{array}\right){x}^{0}{y}^{4}$

$\left(x+y{\right)}^{4}=\frac{4!}{0!4!}{x}^{4}+\frac{4!}{1!3!}{x}^{3}y+\frac{4!}{2!2!}{x}^{2}{y}^{2}+\frac{4!}{3!1!}x{y}^{3}+\frac{4!}{4!0!}{y}^{4}$

$\begin{array}{r}\left(x+y{\right)}^{4}=1{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+1{y}^{4}\\ \left(x+y{\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}$ ### Want to see more solutions like these? 