Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 54740 by gunawan last updated on 10/Feb/19

$$Provethat$$$$1.\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n-1\\ r\end{array}\right)+\left(\begin{array}{c}n-1\\ r-1\end{array}\right)$$$$2.\left(\begin{array}{c}n\\ r\end{array}\right)+\left(\begin{array}{c}n\\ r-1\end{array}\right)=\left(\begin{array}{c}n+1\\ r\end{array}\right)$$$$3.\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+..+\left(\begin{array}{c}n\\ n\end{array}\right)=2^n$$

Answered by Kunal12588 last updated on 10/Feb/19

$$3.{}^n{C}_0{+}^n{C}_1{+}^n{C}_2+...{+}^n{C}_n=2^n{eq}^{n}1$$$$frombinomialexpansion$$$${(a+b)}^p{=}^p{C}_0{a}^p{+}^p{C}_1{a}^{p-1}{b}^1+...{+}^p{C}_r{a}^{p-r}{b}^r+...{+}^p{C}_p{b}^p[r<p]$$$$comparingwiththequestion$$$$\because 1^{anything}=1$$$${eq}^{n}1canbewrittenas:$$$${}^n{C}_0{1}^n{+}^n{C}_1{1}^{n-1}{1}^1{+}^n{C}_2{1}^{n-2}{1}^2+...{+}^n{C}_n{1}^n$$$$={(1+1)}^n=2^{n}proved$$

Commented by gunawan last updated on 10/Feb/19

$$\mathrm{nice}\mathrm{solution}$$$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

$$anotherway{(1+x)}^n={nc}_0+{nc}_{1}x+...+{nc}_n{x}^n$$$$putx=1$$$$2^n={nc}_0+{nc}_1+...+{nc}_n$$

Commented by maxmathsup by imad last updated on 11/Feb/19

$$3)letp\left(x\right)=\sum _{k=0}^n{C}_n^k{x}^k={(x+1)}^n$$$$x=1\Rightarrow \sum _{k=0}^n{C}_n^k=2^n\Rightarrow C_n^0+C_n^1+C_n^2+....+C_n^n=2^n.$$

Answered by Kunal12588 last updated on 10/Feb/19

$${}^n{C}_r={}^{n-1}{C}_r{+}^{n-1}{C}_{r-1}$$$$1.RHS$$$${}^{n-1}{C}_r{+}^{n-1}{C}_{r-1}=\frac{(n-1)!}{r!(n-1-r)!}+\frac{(n-1)!}{(r-1)!(n-r)!}$$$$=\frac{(n-1)!}{r(r-1)!(n-1-r)!}+\frac{(n-1)!}{(r-1)!(n-r)(n-1-r)!}$$$$=\frac{(n-1)!}{(r-1)!(n-1-r)!}(\frac{1}{r}+\frac{1}{n-r})$$$$=\frac{(n-1)!}{(r-1)!(n-1-r)!}\left(\frac{n}{r(n-r)}\right)$$$$=\frac{n(n-1)!}{r(r-1)!(n-r)(n-1-r!)}=\frac{n!}{r!(n-r)!}$$$$={}^n{C}_r=LHSproved$$

Answered by Kunal12588 last updated on 10/Feb/19

$$2.{}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$$$$LHS$$$$=\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$$$$=\frac{n!}{(r-1)!(n-r)!}(\frac{1}{r}+\frac{1}{n-r+1})$$$$=\frac{(n+1)!}{r!(n+1-r)!}{=}^{n+1}{C}_r=RHSproved$$

Commented by Kunal12588 last updated on 10/Feb/19

$$sameas\left(1\right)justreplace(n-1)withn$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com