Question 3, Exercise 6.2

Solutions of Question 3 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 $r$, if: $^6P_{r-1}=^5P_4$

Solution.

$$\begin{align*}{ }^{6} P_{r-1}&={ }^{5} P_{4}\\ \dfrac{6!}{(6-(r-1))!}&=\dfrac{5!}{(5-4)!}\\ \dfrac{6!}{(7-r)!}&=\dfrac{5!}{1!} \\ \frac{6 \times 5!}{(7-r)!}=\dfrac{5!}{1} 6&=(7-r)!\\ 3!&=(7-r)!\\ 3&=7-r \\ r&=7-3\\ r&=4\end{align*}$$

Find $r$, if: $^{10}P_{r}=2\,^9P_r$

Solution.

$$\begin{align*}{ }^{10} P_{r}&=2 \times{ }^{9} P_{r}\\ \dfrac{10!}{(10-r)!}&=2 \cdot \dfrac{9!}{(9-r)!}\\ \dfrac{10 \cdot 9!}{(10-r)!}&=2 \cdot \dfrac{9!}{(9-r)!}\\ 10(9-r)!&=2(10-r)!\\ 10(9-r)!&=2(10-r)(9-r)!\\ 10&=2(10-r)\\ 5&=10-r\\ r&=5 \end{align*}$$

Find $r$, if: $^{15}P_{r}=210$

Solution.

$$\begin{align*}^{15}P_{r}&=210\\ \dfrac{15!}{(15-r)!}&=210\\ (15-r)!&=(15-2)!\\ r&=2\end{align*}$$

Find $r$, if: $^{10}P_{r}=3\,^{10}P_{r-1}$

Solution.

$$\begin{align*}^{10}P_{r}&=3\,^{10}P_{r-1}\\ \dfrac{10!}{(10-r)!}&=3\dfrac{10!}{(10-(r-1))!}\\ \dfrac{1}{(10-r)!}&=3\dfrac{1}{(11-r)!}\\ (11-r)!&=3(10-r)!\\ (11-r)(10-r)!&=3(10-r)!\\ 11-r&=3\\ r&=8\end{align*}$$

Find $r$, if: $4\,^6P_{r}=^6P_{r+1}$

Solution.

$$\begin{align*}4 \cdot{ }^{6} P_{r}&={ }^{6} P_{r+1}\\ 4 \cdot \dfrac{6!}{(6-r)!} & =\dfrac{6!}{(6-(r+1))!} \\ \dfrac{4}{(6-r)!} & =\dfrac{1}{(6-r-1)!} \\ \dfrac{4}{(6-r)!} & =\dfrac{1}{(5-r)!}\\ 4(5-r)! & =(6-r)!\\ 4(5-r)! & =(6-r)(5-r)! \\ 4 & =6-r \\ r & =6-4 \\ r & =2 \end{align*}$$

Find $r$, if: $2 .{ }^{6} P_{r-1}={ }^{5} P_{r}$

Solution.

$$\begin{align*}2 .{ }^{6} P_{r-1}&={ }^{5} P_{r}\\ 2. \dfrac{6!}{(6-(r-1))!}&=\dfrac{5!}{(5-r)!}\\ 2. \dfrac{6!}{(6-r+1)!}&=\dfrac{5!}{(5-r)!} \\ 2 \cdot \dfrac{6 \cdot 5!}{(7-r)!}&=\dfrac{5!}{(5-r)!}\\ 12(5-r)!& =(7-r)! \\ 12(5-r)!& =(7-r)(6-r)(5-r)! \\ 12& =(7-r)(6-r) \\ 12& =42-13 r+r^{2} \\ r^{2}-13 r+30& =0 \\ r^{2}-10 r-3 r+30& =0 \\ r(r-10)-3(r-10)& =0 \\ (r-10)(r-3)& =0 \\ r \text { must be <n }& \quad r=10, r=3 \\ r& =3 \\ \end{align*}$$

Find $r$, if: $^{54} P_{r+3}:^{56} P_{r+6}=1: 30800$

Solution.

$$\begin{align*}{ }^{54} P_{r+3}:^{56} P_{r+6}&=1: 30800 \\ \dfrac{54!}{(54-(r+3))!}: \dfrac{56!}{(56-(r+6))!}&=1: 30800\\ \dfrac{54!}{(54-r-3)!}: \dfrac{56 \times 55 \times 54!}{(56-r-6)}&=1: 30800 \\ \dfrac{1}{(51-r)!}: \dfrac{3080}{(50-r)!}&=1: 30800 \\ \dfrac{1}{(51-r)(50-r)!}: \dfrac{3080}{(50-r)!}&=1: 30800\\ \dfrac{1}{(51-r) 3080}&=\dfrac{1}{30800}\\ 30800 & =3080(51-r) \\ 51-r & =10 \\ r & =51-10\\ r&=41 \end{align*}$$

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