Question 1(vi-x), Exercise 6.3

Solutions of Question 1(vi-x) of Exercise 6.3 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 $n\in N$: $\quad^{2n}C_n=\dfrac{2^n[1.3.5.\cdots(2n-1)]}{n!}$

Solution.

$$\begin{align*}L.H.S &=\quad^{2n}C_n \\ &=\dfrac{(2 n)!}{n!(2 n-n)!}\\ &=\dfrac{(2 n)(2 n-1)(2 n-2)(2 n-n)(2 n-(n+1)) ..2 .1}{n!\cdot n!}\\ &=\dfrac{(2 n)(2 n-1)(2 n-2) ..(n)(n-1)(n-2) ..2 .1}{n!\cdot n!}\\ &=\dfrac{[(2 n)(2 n-2)(2 n-4) \ldots 4 \cdot 2]}{n!n!}\\ &\times \dfrac{[(2 n-1)(2 n-3) \ldots 3.1]}{n!n!}\\ &=\dfrac{2^{n}[n(n-1)(n-2) \ldots 2 \cdot 1]}{n}\\ &\times\dfrac{[(2 n-1) \ldots 3.1]}{n}\\ &=\dfrac{2^{n} \cdot n![(2 n-1) \ldots 3 \cdot 1]}{n!n!}\\ &=\dfrac{2^{n}(1 \cdot 3 \cdot 5 \ldots(2 n-1))}{n!}= R.H.S\end{align*}$$

Prove for $n\in N$: $\quad^nC_p=^nC_q\implies p=q\,\,or\,\,p+q=n$

Solution.

Let $$\begin{align*}{ }^{n} C_{p}&={ }^{n} C_{q}\\ \Rightarrow \dfrac{n!}{p!(n-p)!}&=\dfrac{n!}{q!(n-q)!}\\ q!(n-q)!&=p!(n-p)!\\ \Rightarrow q!&=p!\quad i.e. \quad q=p \quad \\ \text{or} \quad q!&=(n-p)!\\ i.e.\quad q&=n-p\\ n&=p+q\end{align*}$$

Prove for $n\in N$: $\,\,^nC_r+2^nC_{r-1}+^nC_{r-2}=\,^{n+2}C_r$

Solution.

$$\begin{align*}L.H.S& ={ }^{n} C_{p}+2^{n} C_{r-1}+{ }^{n} C_{r-2}={ }^{n+2} C_{r}\\ &= \dfrac{n!}{r!(n-r)!}+\dfrac{2 n!}{(r-1)!(n-(r-1))!}\\ &+\dfrac{n!}{(r-2)!(n-(r-2))!} \\ &= \dfrac{n!}{r!(n-r)!}+\dfrac{2 n!}{(r-1)!(n-r+1)!}\\ &+\dfrac{n!}{(r-2)!(n-r+2)!} \\ &=\dfrac{n!}{r(r-1)(r-2)!(n-r)!}\\ &+\dfrac{2 n!}{(r-1)(r-2)!(n-r+1)(n-r)!}\\ &+\dfrac{n!}{(r-2)!(n-r+2)(n-r+1)(n-r)!}\\ &=\dfrac{n!((n-r+1)+2r)}{r(r-1)(r-2)!(n-r+1)(n-r)!}\\ &+\dfrac{n!}{(r-2)!(n-r+2)(n-r+1)(n-r)!}\\ &=\dfrac{n!(n+r+1)}{r(r-1)(r-2)!(n-r+1)(n-r)!}\\ &+\dfrac{n!}{(r-2)!(n-r+2)(n-r+1)(n-r)!}\\ &=\dfrac{n![(n-r+2)(n+r+1)+r(r-1)]}{r(r-1)(r-2)!(n-r+2)(n-r+1)(n-r)!}\\ &=\dfrac{n![n^2+nr+n-rn-r^2-r+2n+2r+2]}{{r!(n-r+2)!}}\\ &+\dfrac{n![r^2-r]}{{r!(n-r+2)!}}\\ &=\dfrac{n![n^2+3n-r^2+r+2]}{{r!(n-r+2)!}}\\ &+\dfrac{n![r^2-r]}{{r!(n-r+2)!}}\\ &=\dfrac{n![n^2+3n-r^2+r+2+r^2-r]}{{r!(n-r+2)!}}\\ &=\dfrac{n!\left[n^{2}+3 n+2\right]}{r!(n-r+2)!}\\ &=\dfrac{n!\left[n^{2}+n+2 n+2\right]}{r!((n+2)-r)!}\\ &=\dfrac{n![n(n+1)+2(n+1)]}{r!((n+2)-r)!}\\ &=\dfrac{n!(n+1)(n+2)}{r!((n+2)-r)!} \quad \because n!(n+1)(n+2)=(n+2)!\\ &=\dfrac{(n+2)!}{r!((n+2)-r)!}={ }^{n+2} C_{r}= R.H.S\end{align*}$$

Prove for $n\in N$: $\,\,r\cdot^nC_r=\,n\,\,^{n-1}C_{r-1}$

Solution.

$$\begin{align*} L.H.S &=r\cdot^nC_r\\ &=r \cdot \dfrac{n!}{r!(n-r)!} \quad \because r!\geq r(r-1)!\\ & = \dfrac{n!}{(r-1)(n-r)!} \quad \because n!=n(n-1)! \\ & =n \cdot \dfrac{(n-1)!}{(r-1)!((n-1)-(r-1))!} \\ & =n^{n-1} C_{r-1}=\text { R.H.S } \end{align*}$$

Prove for $n\in N$: The product of $k$ consecutive integer is divisible by $k!$.

Solution.

As product of $1^{st} k$ positive integers is

$$1 \times 2 \times 3 \times 4 \times \ldots \times k=k!$$

as this product is equal to $k$ ! so it is divisible by $k$ !

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