Question 7(i-vi), Exercise 6.1

Solutions of Question 7(i-vi) 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.

Find $n$, if $\quad \dfrac{n!}{(n-2)!}=930,\quad n \geq 2$

Solution.

$$\begin{align*} \dfrac{n!}{(n-2)!}&=930\\ \dfrac{n(n-1)(n-2)!}{(n-2)!}&=930\\ n(n-1)&=930\\ n^2-n-930&=0 \end{align*}$$ Byusing quadratic formula, $$\begin{align*} n&=\dfrac{1\pm \sqrt{1+4(930)}}{2}\\ &=\dfrac{1\pm 61}{2}\\ n&=\dfrac{1+61}{2},\quad or \quad n=\dfrac{1-61}{2}\\ n&=31,\quad or \quad n=-30\\ \end{align*}$$ Since $$n\geq 2\implies n=31$$

Find $n$, if $\quad \dfrac{n!}{(n-5)!}=20\cdot \dfrac{n!}{(n-3)!}, \quad n \geq 5$

Solution.

$$\begin{align*} \dfrac{n!}{(n-5)!}&=20\cdot \dfrac{n!}{(n-3)!}\\ \dfrac{n!}{(n-5)!}&=20\cdot \dfrac{n!}{(n-3)(n-4)(n-5)!}\\ (n-3)(n-4)&=20\\ n^2-7n+12&=20\\ n^2-7n-8&=0\\ n^2+n-8n-8&=0\\ n(n+1)-8(n+1)&=0\\ (n+1)(n-8)&=0\\ n+1&=0,\quad or \quad n-8=0\\ n&=-1,\quad or \quad n=8\\ \end{align*}$$ Since $$n \geq 5 \implies n=8$$

Find $n$, if $\quad (n+2)!= 60\cdot(n-1)!$

Solution.

$$\begin{align*} (n+2)!&= 60(n-1)!\\ (n+2)!&= 60(n-1)!\\ (n+2)(n+1)n(n-1)!&= 60(n-1)!\\ (n+2)(n+1)n&= 60\\ (n^2+3n+2)n&= 60\\ n^3+3n^2+2n-60&= 0\\ \end{align*}$$ $$\begin{array}{c|cccc} & 1 & 3 & 2 & -60 \\ 3& 0 & 3 & 18 & 60\\ \hline & 1 & 6 & 20 & 0 \\ \end{array}$$ $\implies n=3$ remaining roots are imaginary.

Find $n$, if $\quad (n+2)!= 132\cdot n!$

Solution.

$$\begin{align*} (n+2)!&= 132\cdot n!\\ (n+2)(n+1)n!&= 132\cdot n!\\ (n+2)(n+1)&= 132\\ (n+2)(n+1)&= 132\\ n^2+3n-130&=0\\ n^2+13n-10-130&=0\\ n(n+13)-10(n+13)&=0\\ (n+13)(n-10)&=0\\ n+13&=0\quad or \quad n-10=0\\ n&=-1 \quad or \quad n=10 \end{align*}$$ $n$ cannot be negative,so $n=10$

Find $n$, if $\quad (n+2)!= 56\cdot n!$

Solution.

$$\begin{align*} (n+2)!&= 56n!\\ (n+2)(n+1)n!&= 56 n!\\ n^2+3n+2&= 56 \\ n^2+3n-54&= 0 \\ n^2+9n-6n-54&= 0 \\ n(n+9)-6(n+9)&= 0 \\ (n+9)(n-6)&= 0 \\ n+9&=0\quad or\quad n-6= 0 \\ n&=-9\quad or\quad n=6 \\ \end{align*}$$ $n$ cannot be negative,so $n=6$

Find $n$, if $\quad \dfrac{1}{9!}+\dfrac{1}{10!}=\dfrac{n}{11!}$

Solution.

$$\begin{align*} \dfrac{1}{9!}+\dfrac{1}{10!}&=\dfrac{n}{11!}\\ \dfrac{10\times 11}{9!10\times 11}+\dfrac{11}{10!\times 11}&=\dfrac{n}{11!}\\ \dfrac{110}{ 11!}+\dfrac{11}{11!}&=\dfrac{n}{11!}\\ \dfrac{110+11}{ 11!}&=\dfrac{n}{11!}\\ 121&=n \end{align*}$$

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