Question 2, Exercise 6.3

Solutions of Question 2 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 $n$, if : $\,\, ^nC_5=\,\, ^nC_8$

Solution.

$$\begin{align*}\dfrac{n!}{5!(n-5)!}=\dfrac{n!}{8!(n-8)!}&\\ \dfrac{1}{8!(n-5)(n-6)(n-7)(n-8)!}&\\ =\dfrac{1}{8 \times 7 \times 6 \times 8!(n-8)!}&\\ 336=(n-5)(n-6)(n-7)&\\ (n-5)\left(n^{2}-13 n+42\right)&=336\\ n^{3}-13 n^{2}+42 n-5 n^{2}+65 n-210&=336\\ n^{3}-18 n^{2}+107 n-546&=0\end{align*}$$ By Synthetic division
$$\begin{array}{c|cccc} & 1 & -18 & 107 & -546 \\ 13 & 0 & 13 & -65 & 546\\ \hline & 1 & -5 & 42 & 0 \\ \end{array}$$ Since remaining roots of $n$ are imaginary so $n=3$.

Find $n$, if : $\,\, ^nC_{15}=\,\, ^nC_7$

Solution.

Since $$\begin{align*}{ }^{n} C_{r}&={ }^{n} C_{n-r}\\ \Rightarrow \quad \text{If} r\quad &=7\quad\text{then}\\ n-r&=15\\ \text{or} \quad n-7&=15 \\ \Rightarrow n &=15+7=22\end{align*}$$

Find $n$, if : $\,\, ^nC_{50}=\,\, ^nC_1$

Solution.

As we know $$\begin{align*}{ }^{n} C_{1}&={ }^{n} C_{n-1}\\ \Rightarrow 50&=n-1\\ \Rightarrow n& =51\end{align*}$$

Find $n$, if : $\,\, ^{2n}C_3:^nC_3=\,\,11:1$

Solution.

$$\begin{align*}\dfrac{(2 n)!}{(2 n-3)!}: \dfrac{n!}{3!(n-3)!}&=11: 1\\ \dfrac{(2 n)(2 n-1)(2 n-2)(2 n-3)!}{(2 n-3)!}&\\ : \dfrac{n(n-1)(n-2)(n-3)!}{(n-3)!}&=11: 1\\ \dfrac{2 n(2 n-1)(2 n-2)}{n(n-1)(n-2)}&=\dfrac{11}{1} \\ \dfrac{2\left(4 n^{2}-4 n-2 n+2\right)}{n^{2}-2 n-n+2}&=11\\ 8n^2-12n+4=11n^2&-33n+22\\ 3n^2-21n+18&=0\\ 3(n^2-7n+6)&=0\\ n^2-7n+6&=0\\ (n-1)(n-6)&=0\quad 3\neq 0\\ n-1=0\quad &\text{or}\quad n-6=0\\ n=1\quad &\text{or}\quad n=6\end{align*}$$ $n$ should be greater than $3$, so $n=6$.

Find $n$, if : $\,\, ^{n}C_6:\,\,^{n-3}C_3=\,\,33:4$

Solution.

$$\begin{align*}\dfrac{n!}{6!(n-6)!}: \dfrac{(n-3)!}{3!((n-3)-3)!}&=33: 4\\ \dfrac{n!}{6 \times 5 \times 4 \times 3!(n-6)!}: \dfrac{(n-3)!}{3!(n-6)!}&=33: 4\\ \dfrac{n(n-1)(n-2)(n-3)!}{120}: \dfrac{(n-3)!}{1}&=33: 4\\ \dfrac{n(n-1)(n-2)}{120}=\dfrac{33}{4}\\ n(n-1)(n-2) & =990 \\ n\left(n^{2}-3 n+2\right) & =990 \\ n^{3}-3 n^{2}+2 n & =990\\ n^{3}-3 n^{2}+2 n -990 &=0\end{align*}$$ By Synthetic division
$$\begin{array}{c|cccc} & 1 & -3 & 2 & -990 \\ 11 & 0 & 11 & 88 & 990\\ \hline & 1 & 8 & 90 & 0 \\ \end{array}$$ Since remaining roots of $n$ are imaginary so $n=11$.

Find $n$, if : $\,\, ^{2n}C_3:^{n}C_2=\,\,12:1$

Solution.

$$\begin{align*}\dfrac{(2 n)!}{3!(2 n-3)!}: \dfrac{n!}{2!(n-2)!}&=12: 1\\ \dfrac{(2 n)(2 n-1)(2 n-2)(2 n-3)!}{3 \cdot 2!(2 n-3)!}&\\ : \dfrac{n(n-1)(n-2)!}{2!(n-2)!}&=12: 1\\ \dfrac{2n(2 n-1)(2 n-2)}{3n(n-1)}&=\dfrac{12}{1}\\ 2(2 n-1)(2 n-2)&=36(n-1) \\ 2\left(4 n^{2}-4 n-2 n+2\right)&=36 n-36 \\ 8 n^{2}-12 n+4&=36 n-36 \\ 8 n^{2}-48 n+40 &=0 \\ 8\left(n^{2}-6 n+5\right)&=0 \\ n^{2}-6 n+5&=0 \\ n^{2}-n-5 n+5&=0\quad\text{where} 8 \neq 0 \\ n(n-1)-5(n-1)&=0 \\ (n-1)(n-5)&=0 \\ n-1&=0 \\ & n=1\\ \text { or } \quad n-5&=0 \\ n&=5 \end{align*}$$ $n$ must be greater or equal to $r=2$, then $n=5$

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