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- Question 2, Exercise 6.3
- e \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*} =====Question 2(iii)====... w \begin{align*}{ }^{n} C_{1}&={ }^{n} C_{n-1}\\ \Rightarrow 50&=n-1\\ \Rightarrow n& =51\end{align*} =====Question 2(iv)===== Find $n$, if : $\,\, ^{2n}C_3:^nC_3=\,\,1
- Question 1(i-v), Exercise 6.3
- \\ $$ \begin{array}{ll} & r!X={ }^{n} P_{r} \\ \Rightarrow & r!X=\frac{n!}{(n-r)!} \\ \Rightarrow & X=\frac{n!}{r!(n-r)!}={ }^{n}{C}_{r} \end{array} $$ =====Question 1(... r)!} \\ & \quad \because(n-r)!(n-r+1)=(n-r+1)!\\ \Rightarrow \quad&=(n-r+1) \frac{n!}{(r-1)!(n-r+1)!}\\ & =(n-
- Question 1(vi-x), Exercise 6.3
- Let \begin{align*}{ }^{n} C_{p}&={ }^{n} C_{q}\\ \Rightarrow \dfrac{n!}{p!(n-p)!}&=\dfrac{n!}{q!(n-q)!}\\ q!(n-q)!&=p!(n-p)!\\ \Rightarrow q!&=p!\quad i.e. \quad q=p \quad \\ \text{or} \qu
- Question 1, Exercise 6.2
- }{(n-n+1)!}\\ &=\dfrac{n!}{1!}=n!\because 1!=1\\ \Rightarrow\quad L.H.S &= R.H.S\end{align*} =====Question 1(
- Question 2, Exercise 6.2
- , \\ & n=8 \end{align*} $n$ must be an integer, $\Rightarrow n=8$ =====Question 2(ix)===== Find $n$, if: $\qu
- Exercise 6.3 (Solutions)
- 2 n-1)]}{n!}$ (vii) ${ }^{n} C_{p}={ }^{n} C_{q} \Rightarrow p=q$ or $p+q=n$ \\ (viii) ${ }^{n} C_{r}+2{ }^{n}