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Question Number 16876 by tawa tawa last updated on 27/Jun/17

$$\mathrm{In}\mathrm{how}\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{a}\mathrm{family}\mathrm{of}5\mathrm{brothers}\mathrm{be}\mathrm{seated}\mathrm{round}\mathrm{a}\mathrm{table}$$$$\mathrm{if}\left(\mathrm{i}\right)2\mathrm{brothers}\mathrm{must}\mathrm{seat}\mathrm{next}\mathrm{to}\mathrm{each}\mathrm{other}.$$$$\left(\mathrm{ii}\right)2\mathrm{brothers}\mathrm{must}\mathrm{not}\mathrm{seat}\mathrm{together}.$$

Answered by mrW1 last updated on 27/Jun/17

$$\left(\mathrm{i}\right)$$$$2\times (4-1)!=12$$$$\left(\mathrm{ii}\right)$$$$(5-1)!-12=12$$

Commented by tawa tawa last updated on 27/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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