Question Number 110157 by I want to learn more last updated on 27/Aug/20
$$\mathrm{If}\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{5}\:\:\mathrm{people},\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{they}\:\mathrm{be}\:\mathrm{seated} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{round}\:\mathrm{table},\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}, \\ $$$$\left(\mathrm{a}\right)\:\:\:\mathrm{7}\:\:\mathrm{chairs}\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\:\:\:\mathrm{3}\:\:\mathrm{chairs}\:\:\:\:\:\:\:\:\:\:\:\mathrm{available} \\ $$
Answered by bemath last updated on 27/Aug/20
$$\left({a}\right)\:{C}_{\mathrm{5}} ^{\mathrm{7}} ×\mathrm{4}!\:=\:\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}×\mathrm{1}}×\mathrm{24}=\mathrm{21}×\mathrm{24}\:{way} \\ $$
Answered by I want to learn more last updated on 27/Aug/20
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$