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Question Number 227098 by gregori last updated on 31/Dec/25
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 31/Dec/25
$${p}=\frac{\mathrm{13}}{{C}_{\mathrm{3}} ^{\mathrm{13}−\mathrm{1}+\mathrm{3}} }=\frac{\mathrm{13}}{\mathrm{455}}=\frac{\mathrm{1}}{\mathrm{35}} \\ $$$${or} \\ $$$${p}=\frac{\mathrm{13}×{C}_{\mathrm{3}} ^{\mathrm{4}} }{{C}_{\mathrm{3}} ^{\mathrm{52}} }=\frac{\mathrm{52}}{\mathrm{22100}}=\frac{\mathrm{1}}{\mathrm{425}} \\ $$
Commented by gregori last updated on 31/Dec/25
$$\:{why}\:{the}\:{answer}\:{is}\:{not}\:{same}? \\ $$
Commented by mr W last updated on 31/Dec/25
$${it}\:{depends}\:{on}\:{whether}\:{the}\:{cards} \\ $$$${with}\:{same}\:{number}\:{are}\:{treated}\:{as} \\ $$$${distinct}\:{or}\:{indistinct}\:{cards}. \\ $$
Commented by gregori last updated on 31/Dec/25
$${the}\:{cards}\:{are}\:{identicall}\:{sir} \\ $$
Commented by mr W last updated on 31/Dec/25
$${then}\:{p}=\frac{\mathrm{1}}{\mathrm{35}} \\ $$