Question

plzz solve one and two.​

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$\sf\textsf{ Find r , if ^{n}_{r} P=720 and ^{n}_{r}C =120}$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\boxed{\begin{minipage}{5 cm}\begin{aligned}\bigstar\: \quad \bf ^{n}_{r}P\quad &= \dfrac{n !}{(n-r)!}\\\atop\\\bigstar\: \quad \bf ^{n}_{r}C \quad &= \dfrac{n !}{(n-r)! \cdot r!}\end{aligned}\end{minipage}}$

$\mathsf{Divide\: ^{n}_{r}P\: by\: ^{n}_{r}C}$

$\quad \mathsf{\dfrac{^n_rP}{^n_rC}=\dfrac{720}{120}}$

$\implies \mathsf{\dfrac{\frac{n!}{(n-r)!}}{\frac{n!}{(n-r)!\cdot r!}} = \dfrac{72\cancel{0}}{12\cancel{0}}}$

$\implies \mathsf{\dfrac{\cancel{n!}}{\cancel{(n-r)!}} \times \dfrac{\cancel{(n-r)!} \cdot r!}{\cancel{n!}} = 6 }$

$\implies \mathsf{r! = 6}$

$\implies \mathsf{r! = 3!}$

$\therefore$

$\blacksquare\: \: \LARGE{\underline{\mathbf{r \: = \: \boxed{\utilde{3}}}}}$



$\underline{\underline{\Huge{\textbf{Question:}}}}$

$\sf\textsf{Prove that \sum^{n}_{r=0}\, 3^r\: ^n_rC\: =\: 4^n }$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\sf\textbf{From Binomial Theorem}$

$\boxed{\begin{minipage}{10 cm} \bigstar \quad \tiny{\tt \sum\limits_{r=0}^{n} \, ^n_rC \: a^{n-r}\: b^r\; = \; ^nC_0\, a^n b^0 + ^nC_1\, a^{n-1}\, b^1+ \cdots \cdots + ^nc_{n-1}\, a^1\, b^{n-1}+^nC_n\, a^n\, b^0}\\\atop\bigstar \quad \bf \sum\limits_{r=0}^{n} \, ^n_rC \: a^{n-r}\: b^r \; = \; (a+b)^{n}\rule{2cm}{\fboxrule}\; [1]\end{minipage}}$

$\sf\textsf{Plug in 1 for a and 3 for b into the eq [1]}$

$\implies \mathsf{\sum\limits_{r=0}^{n} \, ^n_rC \: 1^{n-r}\: 3^r \; = \; (1+3)^{n}}$

$\implies \mathsf{\sum\limits_{r=0}^{n} \, ^n_rC \: (1)\: 3^r \; = \; (4)^{n}}$

$\therefore$

$\blacksquare \: \: \LARGE{\underline{\Large{\mathbf{\sum^{n}_{r=0}\, 3^r\: ^n_rC\: =\: 4^n}}}}$