Question 6(vi-ix), Exercise 6.1

Solutions of Question 6(vi-ix) 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 $n\in N$: $\quad 33!$ is divisible by $2^{15}$.

Solution.

$$33!=33.32.31\cdots4.3.2.1$$ $2,4,8,16,32$ are factors of the expression.
and we can write it as $2^1,2^2,2^3,2^4,2^5$
$$\begin{align*}33!&=2^5.2^4.2^3.2^2.2(33\times31\times\cdots6\times5\times3\times1)\\ &=2^{15}(33\times31\times\cdots\times3\times1)\end{align*}$$ So $33$ is the multiple of $2^{15}$ and therefore is divisible by $2^{15}$

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

Solution.

$$\begin{align*}L.H.S.&=\dfrac{2^n!}{[(n-1)!]^2}\\ &=\dfrac{(2n)(2n-1)(2n-2)\cdots n(n-1)(n-2)\cdots3.2.1}{[(n-1)!]^2}\\ &=\dfrac{(2n)(2n-1)(2n-2)\cdots n(n-1)!}{[(n-1)!]^2}\\ &=\dfrac{(2n)(2n-1)(2n-2)\cdots n}{(n-1)!}\\ &=\dfrac{n(n+1)(n+2)\cdots (2n-1)(2n)}{(n-1)!}\\ &=R.H.S. \end{align*}$$

Prove for $n\in N$: $(n!+1)$ is not divisible by any natural number between $2$ and $n$.

Solution.

We know $$n!=n(n-1)(n-2)\cdots 3.2.1$$ Hence $n!$ is divisible by every number between $1$ and $n$.
$n!$ can also divides by any natural number between $2$ and $n$.
For $(n!+1)$, $1$ is not divisible by any natural number between $2$ and $n$.
So $(n!+1)$ is not divisible by any natural number between $2$ and $n$.

Prove for $(n!)^2\leq n^n\cdot n!<(2n)!$

Solution. Consider
$$\begin{align*}(n!)^2&=n!n!\\ (n!)^2&< n^n n!\quad \cdots(i)\\ &\because n!\leq n^n\\ \end{align*}$$ Also condider
$$\begin{align*} (2n)!&=2n(2n-1)(2n-2)\cdots(n+1)n\cdots3.2.1\\ &=2n(2n-1)(2n-2)\cdots(n+1)n!\\ (2n)!&>n^2n!\cdots (ii)\\ & \because 2n(2n-1)(2n-2)\quad \quad\cdots(n+1)>n^2\\ \end{align*}$$ From (i) and (ii), we have
$$(n!)^2\leq n^2 n!<(2n)!$$

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