Question 4, Exercise 6.3

Solutions of Question 4 of Exercise 6.3 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 $n$ and $r$ if: $\,\,^nC_{r-1}:\,^nC_{r}:\,^nC_{r+1}=6:14:21$

Solution.

$$\begin{align*}\dfrac{n!}{(r-1)!(n-(r-1))!}&: \dfrac{n!}{r!(n-r)!}\\ : \dfrac{n!}{(r+1)!(n-(r+1))!} &= 6:14:21\\ \dfrac{1}{(r-1)!(n-r+1)!}: \dfrac{1}{r!(n-r)!}&\\ : \dfrac{1}{(r+1)!(n-r-1)!}&=6: 14: 21\\ \dfrac{1}{(r-1)!(n-r+1)(n-r)(n - r-1)!}&\\ : \dfrac{1}{r(r-1)!(n-r)(n-r-1)!}&\\ : \dfrac{1}{(r+1)(r)(r-1)!(n-r-1)!}&=6: 14: 21\\ \dfrac{1}{(n-r+1)(n-r)}&: \dfrac{1}{r(n-r)}\\ :\dfrac{1}{(r+1) r}=6: 14: 21\\ \dfrac{1}{(n-r+1)(n-r)}: \dfrac{1}{r(n-r)}&=6: 14\\ \text{and}\\ \dfrac{1}{r(n-r)}: \dfrac{1}{(r+1) r}=14: 21\\ \dfrac{r}{n-r+1}&=\dfrac{6}{14} \\ \text { and } \dfrac{r+1}{n-r}&=\dfrac{14}{21}\\ 7 r&=3(n-r+1)\\ \text{and}\\ 3(r+1)&=2(n-r)\\ 7 r&=3 n-3 r+3\\ \text{and}\\ 3r+3&=2n-2r\\ 10r-3n-3&=0\quad \cdots(i)\\ \text{and}\\ 5r-2n+3&=0\quad \cdots(ii)\end{align*}$$

Solving both equations simultaneously.
Multiply (ii) by 2 and subtract from (i) $$\begin{array}{cccc} 10r&-3n&-3&=0 \\ \mathop+\limits_{-}10r&\mathop-\limits_{+}4n&\mathop+\limits_{-}6&=0 \\ \hline &n&-9 &=0\\ \end{array}$$ $$\begin{align*}n & =9 \quad \text{put in (i)}\\ 10 r-3(9)-3 & =0 \\ 10 r-30 & =0 \\ r & =\dfrac{30}{10}\\ r&=3 \end{align*}$$

Find $n$ and $r$ if: $\,\,^nC_{r-1}:\,^nC_{r}:\,^nC_{r+1}=3:4:5$

Solution.

$$\begin{align*}{ }^{n} C_{r-1}:{ }^{n} C_{r}&:{ }^{n} C_{r+1}=3: 4: 5\\ \dfrac{n!}{(r-1)!(n-(r-1))!}&: \dfrac{n!}{r!(n-r)!}\\ : \dfrac{n!}{(r+1)!(n-(r+1))!}&=3: 4: 5\\ \dfrac{1}{(r-1)!(n-r+1)!}&: \dfrac{1}{r!(n-r)!}\\ : \dfrac{1}{(r+1)!(n-r-1)!}&=3: 4: 5\\ \dfrac{1}{(n-1)!(n-r+1)(n-r)(n-r-1)!}&\\ : \dfrac{1}{r(r-1)!(n-r)(n-1)!}&\\ :\dfrac{1}{(r+1)(r)(r-1)!(n-r-1)^{!}}&=3: 4: 5\\ \dfrac{1}{(n-r+1)(n-r)}&: \dfrac{1}{r(n-r)}\\ : \dfrac{1}{(r+1) r}&=3: 4: 5\\ \dfrac{1}{(n-r+1)(n-r)}: \dfrac{1}{r(n-r)}&=3:4\\ \text{and}\\ \dfrac{1}{r(n-r)}: \dfrac{1}{(r+1)r}&=4:5\\ \dfrac{r}{n-r+1}&=\dfrac{3}{4} \\ \text{and}\\ \dfrac{r+1}{n-r}&=\dfrac{4}{5}\\ 4 r&=3(n-r+1)\\ \text{and}\\ 5(r+1)&=4(n-r)\\ 4 r&=3 n-3 r+3\\ \text{and}\\ 5 r+5&=4 n-4 r\\ 7 r-3 n-3&=0\quad \cdots(i)\\ \text{and}\\ 9 r-4 n+5&=0\quad \cdots (ii)\end{align*}$$ Multiply (i) by 4 and (ii) by 3 and subtract $$\begin{array}{cccc} 28r&-12n&-12&=0 \\ \mathop+\limits_{-}27r&\mathop-\limits_{+}12n&\mathop+\limits_{-}15&=0 \\ \hline &r&-27 &=0\\ \end{array}$$ $$\begin{align*} r&=27\quad \text{put in (i)}\\ 7(27)-3 n-3 & =0 \\ 186-3 n & =0 \\ n & =\dfrac{186}{3}\\ n&=62 \end{align*}$$

Find $n$ and $r$ if: $\,\,^{n+1}C_{r+1}:\,^nC_{r}:\,^{n-1}C_{r-1}=22:12:6$

Solution.

$$\begin{align*}{ }^{n+1} C_{r+1}:{ }^{n} C_{r}{ }^{n-1}&: C_{r-1}=22: 12: 6\\ \dfrac{(n+1)!}{(r+1)!((n+1)-(r+1))!}&: \dfrac{n!}{r!(n-r)!}\\ : \dfrac{(n-1)!}{(r-1)!((n-1)-(r-1))!}&=11: 6: 3\\ \dfrac{(n+1)!}{(r+1)!(n-r)!} & : \dfrac{n!}{r!(n-1)!}\\ : \dfrac{(n-1)!}{(r-1)!(n-r)!}&=11: 6: 3 \\ \dfrac{(n+1) n(n-1)!}{(r+1) r(r-1)!}&: \dfrac{n(n-1)!}{r(r-1)!}\\ : \dfrac{(n-1)!}{(r-1)!}&=11: 6: 3\\ \dfrac{(n+1)r}{(r+1)r}: \dfrac{r}{r}&=11: 6\\ \text{and}\\ \dfrac{n}{r}: \dfrac{1}{1}&=6: 3\\ \dfrac{n+1}{r+1}&=\dfrac{11}{6}\\ \text{and}\\ \dfrac{n}{r}&=\dfrac{6}{3}\\ 6(n+1)&=11(r+1)\\ \text{and}\\ 3 n&=6 r\\ 6 n+6 & =11 r+11 \\ 6 n-11 r-5&=0\quad \cdots (i)\\ \text{and}\\ n & =2 r \quad\cdots (ii)\\ \end{align*}$$ Put (ii) in (i)
$$\begin{align*}6(2r)-11r-5&=0\\ 12r-11r&=5\\ r&=5\quad \text{put in (ii)}\\ n&=2(5)\\ n&=10\end{align*}$$

Find $n$ and $r$ if: $\,\,^nC_{r}:\,^nC_{r+1}:\,^nC_{r+2}=1:2:3$

Solution.

$$\begin{align*}{ }^{n} C_{r}:{ }^{n} C_{r+1}&:{ }^{n} C_{r+2}=1: 2: 3\\ \dfrac{n!}{r!(n-r)!}&: \dfrac{n!}{(r+1)!(n-(r+1))!}\\ : \dfrac{n!}{(r+2)!(n-(r+2))!}&=1: 2: 3\\ \dfrac{1}{r!(n-r)!}&: \dfrac{1}{(r+1)!(n-r-1)!}\\ : \dfrac{1}{(r+2)!(n-r-2)}&=1: 2: 3\\ \dfrac{1}{n(n-r)(n-r-1)(n-r-2)!}& \\ :\dfrac{1}{(r+1)r(n-r-1)(n-r-2)!}&\\ :\dfrac{1}{(r+2)(r+1) r(n-r-2)!}&=1: 2: 3\\ \dfrac{1}{(n-r)(n-r-1)}&: \dfrac{1}{(r+1)(n-r-1)}\\ : \dfrac{1}{(r+2)(r+1)}&=1: 2: 3\\ \dfrac{1}{(n-r)(n-r-1)}: \dfrac{1}{(r+1)(n-r-1)}&=1: 2\\ \text{and}\\ \dfrac{1}{(r+1)(n-r-1)}: \dfrac{1}{(r+1)(r+2)}&=2: 3\\ \dfrac{r+1}{n-r}&=\dfrac{1}{2}\\ \text{and}\\ \dfrac{r+2}{n-r-1}&=\dfrac{2}{3}\\ 2(r+1)&=n-r\\ \text{and}\\ 3(r+2)&=2(n-r-1)\\ 2 r+2&=n-r\\ \text{and}\\ 3 r+6&=2 n-2 r-2\\ 3 r-n+2&=0\quad \cdots (1)\\ \text{and}\\ 5 r-2 n+8&=0\quad \cdots (2)\end{align*}$$ Multiply (1) by 2 and subtract from (2)
$$\begin{array}{cccc} 3r&-n&+2&=0 \\ \mathop+\limits_{-}5r&\mathop-\limits_{+}2n&\mathop+\limits_{-}4&=0 \\ \hline &-r&+4 &=0\\ \end{array}$$ $$\begin{align*}r&=4\quad \text{put in (1)}\\ 3(4)-n+2&=0\\ 14-n&=0\\ n&=14\end{align*}$$

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