Question Number 220627 by Spillover last updated on 17/May/25
Answered by mr W last updated on 17/May/25
$$\left.\mathrm{1}\right)\:{no}\:{egg}\:{in}\:{all}\:{three}\:{days}: \\ $$$$\mathrm{3}!=\mathrm{6}\:{ways} \\ $$$$\left.\mathrm{2}\right)\:{egg}\:{in}\:{one}\:{day}: \\ $$$$\mathrm{3}×\mathrm{3}×\mathrm{3}×\mathrm{2}=\mathrm{54}\:{ways} \\ $$$$\left.\mathrm{3}\right)\:{egg}\:{in}\:{the}\:{first}\:{and}\:{last}\:{day} \\ $$$$\mathrm{3}×\mathrm{2}×\mathrm{3}=\mathrm{18} \\ $$$$ \\ $$$${totally}:\:\mathrm{6}+\mathrm{54}+\mathrm{18}=\mathrm{78}\:{ways}\:\checkmark \\ $$
Commented by Spillover last updated on 17/May/25
$${thank}\:{you}.{but}\:{ans}\:\mathrm{78} \\ $$
Commented by mr W last updated on 17/May/25
$${fixed} \\ $$
Commented by Spillover last updated on 17/May/25
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 17/May/25
Commented by Spillover last updated on 17/May/25
$${Egg}\:{dishes}=\mathrm{3}{e}\left({E}_{\mathrm{1}} {E}_{\mathrm{2}} {E}_{\mathrm{3}} \right) \\ $$$${Fish}\:{dishes}=\mathrm{2}\left({F}_{\mathrm{1}} ,{F}_{\mathrm{2}} \right) \\ $$$${Meat}=\mathrm{1}{M} \\ $$
Answered by Spillover last updated on 17/May/25
Commented by Spillover last updated on 17/May/25
where E for Egg, F for Fish and M for Meat while "/" for "or"