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Found in: Page 440

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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# a) What is Pascal’s triangle? b) How can a row of Pascal’s triangle be produced from the one above it?

(a) A geometric arrangement of the binomial coefficients in a triangle which is based on Pascal's identity $n\ge k,\left(\begin{array}{c}n+1\\ k\end{array}\right)=\left(\begin{array}{c}n\\ k\end{array}\right)+\left(\begin{array}{c}n\\ k-1\end{array}\right)$ is known as Pascal’s triangle.

(b) A row of Pascal’s triangle can be produced from the one above it by using the formula $\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)$ .

See the step by step solution

## Step 1: Concept Introduction

Pascal's triangle is a triangular array of binomial coefficients found in probability theory, combinatorics, and algebra in mathematics.

## Step 2: Pascal’s Triangle

(a)

Pascal's triangle is a geometric arrangement of the binomial coefficients in a triangle based on the principle called Pascal's identity which states that for$\mathrm{n}\ge \mathrm{k},\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)$. The row in the triangle consists of the binomial coefficients –

$\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right),\mathrm{k}=0,1,2,\dots ,\mathrm{n}$

Therefore, Pascal triangle is based on Pascal’s identity $\mathrm{n}\ge \mathrm{k},\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)$ .

## Step 3: Row of Pascal’s Triangle

(b)

The ${\mathrm{k}}^{\mathrm{th}}$ term in the ${\left(\mathrm{n}+1\right)}^{\mathrm{th}}$ row in Pascal's triangle is just the sum of the ${\left(\mathrm{k}-1\right)}^{\mathrm{th}}$ and ${\mathrm{k}}^{\mathrm{th}}$ terms in its above row, i.e., the ${\mathrm{n}}^{\mathrm{th}}$ row. Let ${\mathrm{T}}_{\mathrm{k}}\left(\mathrm{n}\right)$ be the ${\mathrm{k}}^{\mathrm{th}}$ term in the ${\mathrm{n}}^{\mathrm{th}}$ row of Pascal's triangle for $\mathrm{k}=0,1,2,...,\mathrm{n},$ then the ${\left(\mathrm{n}+1\right)}^{\mathrm{th}}$ row can be obtained as –

${T}_{k}\left(n+1\right)={T}_{k-1}\left(n\right)+{T}_{k}\left(n\right)\phantom{\rule{0ex}{0ex}}=\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{c}n\\ k\end{array}\right)\phantom{\rule{0ex}{0ex}}=\left(\begin{array}{c}n+1\\ k\end{array}\right)$

Therefore, the row can be obtained using the formula role="math" localid="1668683483371" $\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)$ .

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