Question 2, Exercise 6.2

Solutions of Question 2 of Exercise 6.2 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: $\quad ^nP_4=20\, ^nP_2$

Solution.

$$\begin{align*} \dfrac{m}{(n-4)!}&=20 \cdot \dfrac{m}{(n-2)!}\\ \dfrac{1}{(n-4)!}&=\dfrac{20}{(n-2)(n-3)(n-4)!}\\ (n-2)(n-3)&=20\\ n^{2}-5 n+6&=20\\ n^{2}-5 n-14&=0\\ n^{2}+2 n-7 n-14&=0\\ n(n+2)-7(n+2)&=0\\ (n+2)(n-7)&=0\\ (n+2)&=0 \quad or \quad n-7=0\\ n&=-2\quad or \quad n=7\end{align*}$$ $n$ should be a positive integer, so $n=7$

Find $n$, if: $\quad ^{2n}P_3=100 \, ^nP_2$

Solution.

$$\begin{align*} \dfrac{(2 n)!}{(2 n-3)!}&=100 \cdot \dfrac{n!}{(n-2)!}\\ \dfrac{(2 n)(2 n-1)(2 n-2)(2 n-3)!}{(2 n-3)!}&=100 \cdot \dfrac{n(n-1)(n-2)!}{(n-2)!}\\ 2 n\left[4 n^{2}-4 n-2 n+2\right]&=100\left[n^{2}-n\right]\\ 8 n^{3}-12 n^{2}+4 n&=100 n^{2}-100 n\\ 2 n^{2}-3 n+1 & =25 n-25 \\ 2 n^{2}-28 n+26 & =0 2\left(n^{2}-14 n+13\right)&=0\\ n^{2}-14 n+13&=0\quad \because 2 \neq 0\\ n^{2}-14 n+13&=0 \\ n^{2}-n-13 n+13&=0 \\ n(n-1)-13(n-1)&=0\\ (n-1)(n-13)&=0\\ n-1&=0 \quad \text { or } \quad n-13=0\\ n&=1 \quad \text { or } \quad n=13\end{align*}$$ Since $n \geq r$ i.e., $n \geq 2$ so $n=13$

Find $n$, if: $\quad16\, ^nP_3=13\, ^{n+1}P_3$

Solution.

$$\begin{align*}16 \dfrac{n!}{(n-3)!}&=13 \dfrac{(n+1)!}{((n+1)-3)!}\\ 16 \dfrac{n!}{(n-3)!}&=13 \dfrac{(n+1) n!}{(n-2)(n-3)!}\\ 16(n-2)&=13(n+1)\\ 16 n-32&=13 n+13\\ 16 n-13 n-32-13&=0\\ 3 n-45&=0\\ n&=15\end{align*}$$

Find $n$, if: $\quad ^nP_5=20, ^nP_3$

Solution.

$$\begin{align*}{ }^{n} P_{5}&=20{ }^{n} P_{3}\\ \dfrac{n!}{(n-5)!}&=20 \dfrac{n!}{(n-3)!} \\ \dfrac{1}{(n-5)!}&=\dfrac{20}{(n-3)(n-4)(n-5)!}\\ (n-3)(n-4) & =20 \\ n^{2}-7 n+12 & =20 \\ n^{2}-7 n-8 & =0 \\ n^{2}+n-8 n-8 & =0 \\ n(n+1)-8(n+1) & =0 \\ (n+1)(n-8) & =0 \\ n+1 & =0 \quad \text { or } \quad n-8=0 \\ n & =-1 \quad \text { or } \quad n=8\\ \end{align*}$$ $n$ must be non-negative so $n=8$

Find $n$, if: $30\, ^nP_6= ^{n+2}P_7$

Solution.

$$\begin{align*}30\, ^nP_6&= ^{n+2}P_7\\ 30 \dfrac{n!}{(n-6)!}&=\dfrac{(n+2)!}{(n-5)!}\\ 30 \dfrac{n!}{(n-6)!}&=\dfrac{(n+2)(n+1) n!}{(n-5)(n-6)!}\\ (n-5) 30&=(n+2)(n+1)\\ 30 n-150&=n^{2}+3 n+2\\ n^{2}-27 n+152&=0\end{align*}$$ By using quadratic formula $$\begin{align*} n & =\dfrac{27 \pm \sqrt{(27)^{2}-4(152)}}{2}\\ &=\dfrac{27 \pm \sqrt{121}}{2} \\ & =\dfrac{27 \pm 11}{2} \\ n& =\dfrac{27+11}{2}\quad \text{or}\quad n=\dfrac{27-11}{2} \\ n& =\dfrac{30}{2} \quad \text{or}\quad n=\dfrac{16}{2} \\ n & =19\quad \text{or}\quad n=8 \\ \end{align*}$$

Find $n$, if: $\quad ^nP_5:\, ^{n-1}P_4=6\,:\,1$

Solution.

$$\begin{align*}{ }^{n} P_{5}:{ }^{n-1} P_{4}&=6: 1\\ \dfrac{n!}{(n-5)!}: \dfrac{(n-1)!}{((n-1)-4)!}&=6: 1\\ \dfrac{n!}{(n-5)!}: \dfrac{(n-1)!}{(n-5)!}&=6: 1\\ \dfrac{n(n-1)!}{(n-5)!}: \dfrac{(n-1)!}{(n-5)!}&=6 : 1\\ n: 1&=6: 1\end{align*}$$

Find $n$, if: $\quad ^nP_4:\, ^{n-1}P_3=9\,:\,1$

Solution.

$$\begin{align*} { }^{n} P_{4}:{ }^{n-1} P_{3}&=9: 1\\ \dfrac{n!}{(n-4)!}: \dfrac{(n-1)!}{((n-1)-3)!}&=9: 1\\ n: 1&=9.1 \\ n&=9 \end{align*}$$

Find $n$, if: $\quad ^{n-1}P_3:\, ^{n+1}P_3=5\,:\,12$

Solution.

$$\begin{align*}{ }^{n-1} P_{3}:^{n+1} P_{3}&=5: 12\\ \dfrac{(n-1)!}{((n-1)-3)!}: \dfrac{(n+1)!}{((n+1)-3)!}&=5: 12\\ \dfrac{(n-1)}{(n-4)!}: \dfrac{(n+1)!}{(n-2)!}&=5: 12\\ \dfrac{(n-1)!}{(n-4)!}: \dfrac{(n+1) n(n-1)!}{(n-2)(n-3)(n-4)!}&=5: 12\\ 1: \dfrac{n(n+1)}{(n-2)(n-3)}&=5 \cdot 12\\ \dfrac{1}{\dfrac{n(n+1)}{(n-2)(n-3)}}&=\dfrac{5}{12} \\ \dfrac{(n-2)(n-3)}{n(n+1)}&=\dfrac{5}{12}\\ 12(n-2)(n-3) & =5 n(n+1) \\ 12\left(n^{2}-5 n+6\right) & =5\left(n^{2}+n\right) \\ 12 n^{2}-60 n+72 & =5 n^{2}+5 n \\ 7 n^{2}-65 n+72 & =0 \end{align*}$$ By using quadratic formula $$\begin{align*} & n=\frac{65 \pm \sqrt{(65)^{2}-4(7)(72)}}{14} \\ & n=\frac{65 \pm \sqrt{2209}}{14}=\frac{65 \pm 47}{14} \\ & n=\frac{65+47}{14}, \quad n=\frac{65-47}{14} \\ & n=\frac{112}{14}, \\ & n=8 \end{align*}$$ $n$ must be an integer, $\Rightarrow n=8$

Find $n$, if: $\quad ^{2n-1}P_n:\, ^{2n+1}P_{n-1}=22\,:\,7$

Solution.

$$\begin{align*}{ }^{2 n-1} P_{n}:{ }^{2 n+1} P_{n-1}&=22: 7\\ \dfrac{(2 n-1)!}{((2 n-1)-n)!}: \dfrac{(2 n+1)!}{((2 n+1)-(n-1))!}&=22: 7\\ \dfrac{(2 n-1)!}{(n-1)!}: \dfrac{(2 n+1)!}{(n+2)!}&=22: 7\\ \dfrac{(2 n-1)!}{(n-1)!}: \dfrac{(2 n+1) 2 n(2 n-1)!}{(n+2)(n+1) n(n-1)!}&=22: 7\\ 1: \dfrac{(2n-1)!}{(n-1)!}:\dfrac{(2 n+1)2n(2n-1)}{(n+2)(n+1)n(n+1)!}&=22: 7\\ \dfrac{1}{\dfrac{2(2 n+1)}{(n+1)(n+2)}}&=\dfrac{22}{7}\\ \dfrac{(n+1)(n+2)}{2(2 n+1)}&=\dfrac{22}{7}\\ 7(n+1)(n+2) & =44(2 n+1) \\ 7\left(n^{2}+3 n+2\right) & =88 n+44 \\ 7 n^{2}+21 n+14 & =88 n+44 \\ 7 n^{2}-67 n-30 & =0 \end{align*}$$ By quadratic formula $$\begin{align*} n & =\dfrac{67 \pm \sqrt{67^{2}-4(7)(-30)}}{14} \\ & =\dfrac{67 \pm \sqrt{5329}}{14}=\dfrac{67 \pm 72}{14} \\ n & =\dfrac{67+72}{14},\quad \text{or}\quad n= \dfrac{67-72}{14} \\ n & =\dfrac{140}{14},,\quad \text{or}\quad n=-\dfrac{5}{14} \\ n & =10, \dfrac{-5}{14} \\ \end{align*}$$ $n$ must be a positive integer so $n=10$

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