Question 3, Exercise 6.3

Solutions of Question 3 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 $r$ if : $\quad^{15}C_{3r}= ^{15}C_{3+r}$

Solution.

As we know $${ }^{n} C_{r}={ }^{n} C_{n-r}$$
Here $$\begin{align*}n=15\quad&\text{put} \quad r=3 r\\ n-3 r&=r+3\\ 15-3 r&=r+3 \\ 15-3 & =3 r+r \\ 12 & =4 r\\ r&=3\end{align*}$$

Find $r$ if : $\quad^{8}C_{r}-\,^7C_3= ^{7}C_{2}$

Solution.

$$\begin{align*}^{8}C_{r}-\,^7C_3&= ^{7}C_{2}\\ { }^{7} C_{3}+{ }^{7} C_{2}&={ }^{8} C_{r}\\ { }^{n} C_{r}+{ }^{n} C_{r-1}&={ }^{n+1} C_{r}\end{align*}$$ By comparison, we have $$\begin{align*}n=7,\quad &n+1=8,\\ r=3 \quad &\text { and }\quad r-1=2\\ r=3\end{align*}$$

Find $r$ if : $\quad^{16}C_{r}= ^{16}C_{r+4}$

Solution.

$$\begin{align*}^{16}C_{r}&= ^{16}C_{r+4}\\ \because{ }^{n} C_{r}={ }^{n} C_{n-r}\\ n=16\quad \text{and}& \quad n-r=r+4\\ 16-r&=r+4\\16-4&=r+r\\ 12&=2r\\ r&=6\end{align*}$$

Find $r$ if : $\quad^{15}C_{r}:^{15}C_{r-1}= 11:5$

Solution. $$\begin{align*}^{15}C_{r}:^{15}C_{r-1}&= 11:5\\ \dfrac{(15)!}{r!(15-r)!}: \dfrac{(15)!}{(r-1)!(5-(r-1))!}&=11: 5\\ \dfrac{(15)!}{r!(15-r)!}: \dfrac{(15)!}{(r-1)!(16-r)!}&=11: 5\\ \dfrac{1}{r(r-1)!(15-r)!}: \dfrac{1}{(r-1)!(16-r)(15-r)!}&=11: 5\\ \dfrac{1}{r}: \dfrac{1}{16-r}=11: 5\\ \dfrac{16-r}{r}&=\dfrac{11}{5}\\ 5(16-r) & =11 r \\ 80-5 r & =11 r \\ 80 & =16 r \\ r & =5 \end{align*}$$

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