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Q26E

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Discrete Mathematics and its Applications
Found in: Page 422
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

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

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Short Answer

Let\(n\)and \(k\) be integers with \(1 \le k \le n\). Show that

\(\sum\limits_{k = 1}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} \left( {\begin{array}{*{20}{c}}n\\{k - 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2 - \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right)\)

The expression\(\sum\limits_{k = 1}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} \left( {\begin{array}{*{20}{c}}n\\{k - 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2 - \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right)\)is proved.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Formula of Pascal identity

Pascal identity:

\(\left( {\begin{array}{*{20}{c}}{n + 1}\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}n\\{k - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)\)

Step 2: Calculate the coefficient of \({x^{n + 1}}\) in \({(1 + x)^{2n}}\) in two ways

Coefficient of \({x^k}\) in \({(1 + x)^n}\) multiplied by coefficient of \({(1 + x)^{n - k + 1}}\) in\({(1 + x)^n}\)over all possible values of \(k = \sum\limits_{k = 1}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} \cdot \left( {\begin{array}{*{20}{c}}n\\{n - k + 1}\end{array}} \right)\).

Now, \(k = \sum\limits_{k = 1}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} \cdot \left( {\begin{array}{*{20}{l}}n\\{k - 1}\end{array}} \right)\).

Coefficient of\({x^{n + 1}}\)in\({(1 + x)^{2n}}\)is \(\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right)\).

Step 3: Equate the coefficients and use the Pascal identity to prove the expression

Let \({\bf{n}}\) be a positive integer.

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 1}\\{n + 1}\end{array}} \right)\end{array}\)

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 1)!}}{{(n + 1)!(2n + 1 - (n + 1))!}}\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 1)!}}{{(n + 1)!n!}}\end{array}\)

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 1)!}}{{(n + 1)!n!}} \cdot \frac{{2n + 2}}{{2n + 2}}\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 2)!}}{{(n + 1)!n!}} \cdot \frac{1}{{2n + 2}}\end{array}\)

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 2)!}}{{(n + 1)!n!}} \cdot \frac{1}{{2(n + 1)}}\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 2)!}}{{(n + 1)!(n + 1)!}} \cdot \frac{1}{2}\end{array}\)

\(\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2\)

Hence, \(\sum\limits_{k = 1}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} \left( {\begin{array}{*{20}{c}}n\\{k - 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2 - \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right)\).

Most popular questions for Math Textbooks

a) How many ways are there to deal hands of five cards to six players from a standard 52- card deck?

b) How many ways are there to distribute n distinguishable objects into k distinguishable boxes so that ni objects are placed in box i ?

Prove the identity\(\left( {\begin{array}{*{20}{l}}n\\r\end{array}} \right)\left( {\begin{array}{*{20}{l}}r\\k\end{array}} \right) = \left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)\left( {\begin{array}{*{20}{l}}{n - k}\\{r - k}\end{array}} \right)\), whenever\(n\),\(r\), and\(k\)are nonnegative integers with\(r \le n\)and\(k{\rm{ }} \le {\rm{ }}r\),

a) using a combinatorial argument.

b) using an argument based on the formula for the number of \(r\)-combinations of a set with\(n\)elements.

13. A book publisher has 3000 copies of a discrete mathematics book. How many ways are there to store these books in their three warehouses if the copies of the book are indistinguishable?

11. How many ways are there to choose eight coins from a piggy bank containing 100 identical pennies and 80 identical nickels?

How many permutations of the letters \(ABCDEFGH\) contain

a) the string \(ED\)?

b) the string \(CDE\)?

c) the strings \(BA\) and \(FGH\)?

d) the strings \(AB\;,\;DE\) and \(GH\)?

e) the strings \(CAB\) and \(BED\)?

f) the strings \(BCA\) and \(ABF\)?
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