Question 6(i-v), Exercise 6.1

Solutions of Question 6(i-v) 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 for $n\in N$: $\quad (2n)!=2^n(n!)[1\cdot3\cdot5 \cdots (2n-1)]$

Solution.

$$\begin{align*} (2n)!&= (2n)(2n-1)(2n-2)(2n-3)\cdots (2n-n)(2n-(n+1))\cdots4.3.2.1\\ &= (2n)(2n-1)(2n-2)(2n-3)\cdots (n)(n-1)\cdots4.3.2.1\\ &= (2n)(2n-1)(2n-2)(2n-3)\cdots (n)(n-1)\cdots4.3.2.1\\ &= [(2n)(2n-2)(2n-4)\cdots 6.4.2][(2n-1)(2n-3)\cdots5.3.1]\\ &= [2^n(n)(n-1)(n-2)\cdots 3.2.1][1.3.5.\cdots(2n-1)]\\ &= 2^nn!(1.3.5.\cdots(2n-1))\\ &=R.H.S. \end{align*}$$

Prove for $n\in N$: $\quad (n+1)[n!n+(n-1)!(2n-1)+(n-2)!(n-1)]=(n+2)!$

problem in third term

Solution. $$\begin{align*}L.H.S.&= (n+1)[n!n+(n-1)!(2n-1)+(n-2)!(n-1)!]\end{align*}$$ Multiply and divid 2nd and 3rdterm by $n$. $$\begin{align*} &=(n+1)[n!n+\dfrac{n(n-1)!(2n-1)}{n}+\dfrac{n(n-1)(n-2)!}{n}]\\ &=(n+1)[n!n+\dfrac{n !(2n-1)}{n}+\dfrac{n!}{n}]\\ &=(n+1)n![n+\dfrac{(2n-1)}{n}+\dfrac{1}{n}]\\ &=(n+1)n![n+\dfrac{(2n-1+1)}{n}]\\ &=(n+1)n![n+\dfrac{(2n)}{n}]\\ &=(n+1)!(n+2)\\ &=(n+2)!\\ &=R.H.S. \end{align*}$$

Prove for $n\in N$: $\quad \dfrac{n!}{r!(n-r)!}+\dfrac{n!}{r!(n-r+1)!}=\dfrac{(n+1)!}{r!(n-r+1)!}$

Solution.

Prove for $n\in N$: $\quad \dfrac{n!}{r!}=n(n-1) (n-2)\cdots (r+1)$

Solution.

Prove for $n\in N$: $\quad (n-r+1)\cdot \dfrac{n!}{(n-r+1)!}=\dfrac{n!}{(n-r)!}$

Solution.

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