Question 5 Exercise 6.1

Solutions of Question 5 of Exercise 6.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Show that: $\dfrac{(2 n) !}{n !}=2^n(1.3 .5 \ldots(2 n-1))$

We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(2 n-1)(2 n-2) \\ &=(2 n-3)(2 n-4)(2 n-5) \ldots(2 n-(2 n-4))\\ &(2 n-(2 n-3))(2 n-(2 n-2))(2 n-(2 n-1))]\end{align} In the L.H.S of the above equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(2 n-1)(2 n-2)(2 n-3)(2 n-4)(2 n-5) \ldots 4.3 .2 .1] \\ \dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[2 n(2 n-4)(2 n-6) \ldots 4.2] {[(2 n-1)(2 n-3)(2 n-5) \ldots 5.3 .1]} \end{align} Each underlined brackets contain $n$ terms, that can be simplify as: \begin{align}\dfrac{(2 n) !}{n !}& =\dfrac{1}{n !}[2.2 .2 \ldots 2(n \cdot(n-1)(n-2) \ldots 3.2 .1)] \\ & \times[1.3 .5 \ldots(2 n-3)(2 n-1)] \\ &=\dfrac{1}{n !}2^nn!(1.3 .5 \ldots(2 n-1))\\ \dfrac{(2 n) !}{n !}&=2^n(1.3 .5 \ldots(2 n-1))\end{align}

Show that: $\dfrac{(2 n+1) !}{n !}=2^n(1.3 \cdot 5 \ldots(2 n-1)(2 n+1))$

We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n+1) !}{n !}&=\dfrac{1}{n !}[(2 n+1)(2 n+1-1)(2 n+1-2)\\ &(2 n+1-3)(2 n+1-4)(2 n+1-5)(2 n+1-6) \ldots\\ &(2 n+1-(2 n-2))(2 n+1-(2 n-1))(2 n+1-(2 n-2))(2 n+1-(2 n))]\end{align} In the L.H.S of the above equation. are total $2 n+1$ terms \begin{align}\dfrac{(2 n+1) !}{n !}& =\dfrac{1}{n !}[2 n(2 n-2)(2 n-4)(2 n-6) \\ & \ldots 3.2 .1][(2 n+1)(2 n-1) \ldots 5.3 .1] \\ & =\dfrac{1}{n !}[2.2 .2 \ldots .2(n(n-1) \ldots .3 .2 .1)]\\ & {[1.3 \cdot 5 \ldots \ldots(2 n-1)(2 n+1)]} \\ & =\dfrac{2^n \cdot n !}{n !}[1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)(2 n+1)] \\ & =2^n \cdot[1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)(2 n+1)] \end{align}

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