Question 7(vii-xi), Exercise 6.1

Solutions of Question 7(vii-xi) 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 $\quad n!=990 \cdot (n-3)!$

Solution.

\begin{align*} n!&=990 (n-3)!\\ n(n-1)(n-1)(n-3)!&=990 (n-3)!\\ n(n-1)(n-1)&=990 \\ n^3-3n^2+2n-990 &=0\\ \end{align*} By synhetic divison
\[\begin{array}{c|cccc} & 1 & -3 & 2 & -990 \\ 11 & 0 & 11 & 88 & 990 \\ \hline & 1 & 8 & 90 & 0 \\ \end{array} \] $n=11$ as remaining roots are imaginary.

Find $n$, if $\quad (n+1)!=6 \cdot (n-1)!$

Solution.

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

Find $n$, if $\quad \dfrac{(n+2)!}{(2n-1)!}=\dfrac{(n+3)!}{(2n+1)!}\dfrac{72}{7}$

Solution.

\begin{align*} \dfrac{(n+2)!}{(2n-1)!}&=\dfrac{(n+3)!}{(2n+1)!}\dfrac{72}{7}\\ \dfrac{(n+2)!}{(2n-1)!}&=\dfrac{(n+3)(n+2)!}{(2n+1)2n(2n-1)!}\dfrac{72}{7}\\ 7(2n+1)2n&=72(n+3)\\ 28n^2+14n&= 72n+216\\ 28n^2-58n-216&=0\\ 2(14n^2-29n-108)&=0\\ 14n^2-29n-108&=0\quad 2\neq 0\\ \end{align*} By using quadratic equation,
\begin{align*} n&=\dfrac{29\pm \sqrt{841+6048}}{28}\\ &=\dfrac{29\pm83}{28}\\ n&=\dfrac{29+83}{28}\quad or \quad \dfrac{29-83}{28}\\ n&=\dfrac{112}{28}\quad or \quad \dfrac{-54}{28}\\ n&=4\quad or \quad \dfrac{-27}{14}\\ \end{align*} $n$ cannot be negative, so $n=4$

Find $n$, if $\quad \dfrac{(2n)!}{4!(2n-3)!}\cdot \dfrac{4!(n-4)!}{n!}=52$

Solution.

\begin{align*} \dfrac{(2n)!}{4!(2n-3)!} \dfrac{4!(n-4)!}{n!}&=52\\ \dfrac{(2n)(2n-1)(2n-2)(2n-3)!}{4!(2n-3)!}&\\ \times \dfrac{4!(n-4)!}{n(n-1)(n-2)(n-3)(n-4)!}&=52\\ \dfrac{2(2n-1)(2n-2)}{(n-1)(n-2)(n-3)}&=52\\ 2(2n^2-3n+1)=52(n^3-6n^2+11n-6)&\\ 2n^2-3n+1=13(n^3-6n^2+11n-6)&\\ 2n^2-3n+1=13n^3-78n^2+143n-78)&\\ 13n^3-80n^2+146n-79&=0\\ \end{align*} By synthetic division
\[\begin{array}{c|cccc} & 13 & -80 & 146 & -79 \\ 1& 0 & 13 & -67 & 79 \\ \hline & 13 & -67 & 79 & 0 \\ \end{array} \] $\implies n=1$

Find $n$, if $\quad \dfrac{1}{2(n-2)!}: \dfrac{1}{4!(n-4)!}=2,\quad n \geq 4$

Solution.

\begin{align*} \dfrac{1}{2(n-2)!}: \dfrac{1}{4!(n-4)!}&=2\\ \dfrac{1}{2(n-2)(n-3)(n-4)!}: \dfrac{1}{4!(n-4)!}&=2\\ \dfrac{4!}{2(n-2)(n-3)}&=2\\ 24&=4(n^2-5n+6)\\ 6&=n^2-5n+6\\ n^2-5n&=0\\ n(n-5)&=0\\ n&=5\quad \because n\neq 0\end{align*}

< Question 7(i-iv)