Question Number 49034 by Tinkutara last updated on 01/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
![first man may get Rs 4 or 8 or12...4n−4 [4+(y−1)4=4n−4] [4+4y−4=4n−4 y=n−1 ways] similarly second man (n−1)ways so the required ans is [(n−1+1)(n−1+1)...upto ntimes]−1 n^n −1](https://www.tinkutara.com/question/Q49068.png)
$${first}\:{man}\:{may}\:{get}\:{Rs}\:\mathrm{4}\:{or}\:\mathrm{8}\:{or}\mathrm{12}…\mathrm{4}{n}−\mathrm{4} \\ $$$$\left[\mathrm{4}+\left({y}−\mathrm{1}\right)\mathrm{4}=\mathrm{4}{n}−\mathrm{4}\right] \\ $$$$\left[\mathrm{4}+\mathrm{4}{y}−\mathrm{4}=\mathrm{4}{n}−\mathrm{4}\:\:\:\:{y}={n}−\mathrm{1}\:{ways}\right] \\ $$$${similarly}\:{second}\:{man}\:\left({n}−\mathrm{1}\right){ways} \\ $$$${so}\:{the}\:{required}\:{ans}\:{is}\: \\ $$$$\left[\left({n}−\mathrm{1}+\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)…{upto}\:{ntimes}\right]−\mathrm{1} \\ $$$${n}^{{n}} −\mathrm{1} \\ $$
Commented by Tinkutara last updated on 02/Dec/18
But answer is wrong sir.
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
$${ok}\:{let}\:{me}\:{thik}\:{again}… \\ $$