Question 3 and 4, Exercise 6.1

Solutions of Question 3 and 4 of Exercise 6.1 of Unit 06: Permutation and Combination. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Prove that: $\quad \dfrac{1}{5!}+\dfrac{3}{6!}+\dfrac{1}{7!}=\dfrac{4}{315}$

Solution. $$\begin{align*} LHS = & \dfrac{1}{5!}+\dfrac{3}{6!}+\dfrac{1}{7!} \\ = & \dfrac{1}{5!}+\dfrac{3}{6\cdot 5!}+\dfrac{1}{7\cdot 6\cdot 5!}\\ = & \dfrac{1}{5!}\left(1+\dfrac{3}{6}+\dfrac{1}{7\cdot 6}\right)\\ = & \dfrac{1}{120}\left(1+\dfrac{1}{2}+\dfrac{1}{42}\right)\\ = & \dfrac{1}{120}\cdot \dfrac{32}{21}\\ = & \dfrac{4}{315} = RHS \end{align*}$$

Prove that: $\quad \dfrac{(n-1)!}{(n-3)!}=n^2-3n+2$

Solution. $$\begin{align*} LHS = & \dfrac{(n-1)!}{(n-3)!} \\ = & \dfrac{(n-1)(n-2)(n-3!}{(n-3)!} \\ = & n^2-n-2n+2 \\ = & n^2-3n+2 \\ = & RHS \end{align*}$$

Show that: $\quad \dfrac{(2n)!}{n!}=2^n(1\cdot3\cdot5\cdots(2n-1))$

Solution.

$$\begin{align*} L.H.S.&=\dfrac{(2n)!}{n!}\\ &=\dfrac{2n(2n-1)(2n-2)\cdots (2n-n)(2n-(n+1))\cdots(2n-(2n-1))}{n!}\\ &=\dfrac{2n(2n-1)(2n-2)\cdots (n)(n-1)(n-2)\cdots(2)(1)}{n!}\\ &=\dfrac{[2n(2n-2)\cdots2] [(2n-1)(2n-3)\cdots(3)(1)]}{n!}\\ &=\dfrac{[2n\dot 2(n-1)\cdots2(1)] [(2n-1)(2n-3)\cdots(3)(1)]}{n!}\\ &=\dfrac{2^n[n\dot (n-1)\cdots(1)] [(2n-1)(2n-3)\cdots(3)(1)]}{n!}\\ &=\dfrac{2^nn![(2n-1)(2n-3)\cdots(3)(1)]}{n!}\\ &=2^n[(2n-1)(2n-3)\cdots(3)(1)\\ &= R.H.S. \end{align*}$$

Show that: $\quad \dfrac{(2n-1)!}{n!}=2^{n-1}(1\cdot3\cdot5\cdots(2n-1))$

Solution.

$$\begin{align*} L.H.S.&= \dfrac{(2n-1)!}{n!}\\ &= \dfrac{(2n-1)(2n-2)(2n-3)\cdots(2n-n)(2n-(n+1))(2n-(2n-n))(2n-(2n-1))}{n!}\\ &= \dfrac{(2n-1)(2n-2)(2n-3)\cdots(n)(n-1)(n-2)\cdots (4)(3)(2)(1)}{n!}\\ &= \dfrac{[(2n-2)(2n-4)(2n-6)\cdots(6)(4)(2)][(2n-1)(2n-3)\cdots (5)(3)(1)}{n!}\\ &= \dfrac{2^{n-1}[(n-1)(n-2)\cdots(3)(2)(1)][(2n-1)(2n-3)\cdots (5)(3)(1)}{n!}\\ &= \dfrac{2^{n-1}n!(1.3.5.\cdots(2n-3)(2n-1))}{n!}\\ &=2^{n-1}(1.3.5\cdots(2n-1))\\ &= R.H.S.\end{align*}$$

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