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Question-220874

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Question Number 220874 by Spillover last updated on 20/May/25
Commented by Spillover last updated on 20/May/25
Commented by Spillover last updated on 20/May/25
its n! not n
$${its}\:{n}!\:\:{not}\:{n} \\ $$
Answered by Rasheed.Sindhi last updated on 21/May/25
((n!)/((n−r)!r!))+((n!)/((n−r+1)!(r−1)!)) =((n!)/((n−r)! ×r×(r−1)!))+((n!)/((n−r+1)(n−r)!(r−1)!)) =((n!)/((n−r)! (r−1)!))((1/r)+(1/(n−r+1))) =((n!)/((n−r)! (r−1)!))(((n−r+1+r)/(r(n−r+1)))) =((n!)/((n−r)! (r−1)!))(((n+1)/(r(n−r+1)))) =(((n+1)!)/(r!(n−r+1)!)) =(((n+1)!)/((n+1−r)!r!)) = (((n+1)),(( r)) )
$$\frac{{n}!}{\left({n}−{r}\right)!{r}!}+\frac{{n}!}{\left({n}−{r}+\mathrm{1}\right)!\left({r}−\mathrm{1}\right)!} \\ $$$$=\frac{{n}!}{\left({n}−{r}\right)!\:×{r}×\left({r}−\mathrm{1}\right)!}+\frac{{n}!}{\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!\left({r}−\mathrm{1}\right)!} \\ $$$$=\frac{{n}!}{\left({n}−{r}\right)!\:\left({r}−\mathrm{1}\right)!}\left(\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{n}−{r}+\mathrm{1}}\right) \\ $$$$=\frac{{n}!}{\left({n}−{r}\right)!\:\left({r}−\mathrm{1}\right)!}\left(\frac{{n}−{r}+\mathrm{1}+{r}}{{r}\left({n}−{r}+\mathrm{1}\right)}\right) \\ $$$$=\frac{{n}!}{\left({n}−{r}\right)!\:\left({r}−\mathrm{1}\right)!}\left(\frac{{n}+\mathrm{1}}{{r}\left({n}−{r}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}−{r}+\mathrm{1}\right)!} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}−{r}\right)!{r}!} \\ $$$$=\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix} \\ $$

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