Question Number 142691 by SLVR last updated on 04/Jun/21
$${The}\:{number}\:{of}\:{distributions}\:{of}\:\mathrm{52} \\ $$$${cards}\:{divided}\:{equally}\:{to}\:\mathrm{4}\:{persons}\:{so} \\ $$$${as}\:{each}\:{gets}\:\mathrm{4}\:{cards}\:{of}\:{same}\:{suit} \\ $$$${taken}\:{away}\:{from}\:\mathrm{3}{suits}\left(\mathrm{4}×\mathrm{3}=\mathrm{12}\right)\wp \\ $$$${remaining}\:{card}\:{from}\:{remaining} \\ $$$$\mathrm{4}\:{th}\:{suit}\:{is} \\ $$
Commented by SLVR last updated on 04/Jun/21
$${kindly}\:{help}\:{me}\:{sir}..{I}\:{requested}\: \\ $$$${earlier}..{too}.. \\ $$
Answered by mr W last updated on 17/Jun/21
$${let}\:{A}\:{represent}\:{the}\:\clubsuit\:{cards},\: \\ $$$${B}\:{the}\:\diamondsuit\:{cards}, \\ $$$${C}\:{the}\:\heartsuit\:{cards}\:{and} \\ $$$${D}\:{the}\:\spadesuit\:{cards}. \\ $$$${to}\:{divide}\:{the}\:\mathrm{13}\:{A}−{cards}\:{in}\:{four} \\ $$$${portions}\:{with}\:\mathrm{4},\:\mathrm{4},\:\mathrm{4}\:{and}\:\mathrm{1}\:{cards} \\ $$$${respectively},\:{there}\:{are}\: \\ $$$$\frac{\mathrm{13}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{1}!}=\mathrm{450450}\:{ways}. \\ $$$${to}\:{divide}\:{all}\:{four}\:{suits}\:{there}\:{are}\:{then} \\ $$$$\mathrm{450450}^{\mathrm{4}} \:{ways}. \\ $$$${let}'{s}\:{say}\:{the}\:{portions}\:{of}\:{A}−{cards}\:{are} \\ $$$${A}_{\mathrm{1}} ,{A}_{\mathrm{2}} ,{A}_{\mathrm{3}} \:{and}\:{A}_{\mathrm{4}} .\:{portion}\:{A}_{\mathrm{4}} \:{has}\:{only} \\ $$$${one}\:{card},\:{other}\:{portions}\:{have}\:\mathrm{4}\:{cards}\: \\ $$$${each}. \\ $$$${we}\:{have}\:{now}\:{following}\:{portions}: \\ $$$${A}_{\mathrm{1}} ,{A}_{\mathrm{2}} ,{A}_{\mathrm{3}} ,{A}_{\mathrm{4}} \\ $$$${B}_{\mathrm{1}} ,{B}_{\mathrm{2}} ,{B}_{\mathrm{3}} ,{B}_{\mathrm{4}} \\ $$$${C}_{\mathrm{1}} ,{C}_{\mathrm{2}} ,{C}_{\mathrm{3}} ,{C}_{\mathrm{4}} \\ $$$${D}_{\mathrm{1}} ,{D}_{\mathrm{2}} ,{D}_{\mathrm{3}} ,{D}_{\mathrm{4}} \\ $$$${now}\:{we}\:{distribute}\:{the}\:{A}_{\mathrm{4}} ,{B}_{\mathrm{4}} ,{C}_{\mathrm{4}} ,{D}_{\mathrm{4}} \\ $$$${among}\:{the}\:{four}\:{persons}.\:{there}\:{are} \\ $$$$\mathrm{4}!\:{ways}. \\ $$$${to}\:{distribute}\:{A}_{\mathrm{1}} ,\:{A}_{\mathrm{2}} ,\:{A}_{\mathrm{3}} \:{to}\:{the}\:{three}\: \\ $$$${persons}\:{without}\:{A}_{\mathrm{4}} \:{there}\:{are}\:\mathrm{3}!\:{ways}. \\ $$$${totally}\:{there}\:{are}\:\left(\mathrm{3}!\right)^{\mathrm{4}} \:{ways}. \\ $$$$ \\ $$$${therefore}\:{we}\:{have}\:{totally} \\ $$$$\left(\frac{\mathrm{13}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{1}!}\right)^{\mathrm{4}} ×\mathrm{4}!×\left(\mathrm{3}!\right)^{\mathrm{4}} \approx\mathrm{1}.\mathrm{28}×\mathrm{10}^{\mathrm{27}} \:{ways} \\ $$$$ \\ $$$${to}\:{distribute}\:\mathrm{52}\:{cards}\:{to}\:{the}\:{four} \\ $$$${persons}\:{there}\:{are}\:{totally} \\ $$$$\frac{\mathrm{52}!}{\left(\mathrm{13}!\right)^{\mathrm{4}} }\approx\mathrm{5}.\mathrm{364}×\mathrm{10}^{\mathrm{28}} \:{ways}. \\ $$$${p}\approx\frac{\mathrm{1}.\mathrm{28}}{\mathrm{53}.\mathrm{64}}=\mathrm{2}.\mathrm{38\%} \\ $$
Commented by mr W last updated on 17/Jun/21
$${this}\:{is}\:{my}\:{final}\:{solution}. \\ $$$${do}\:{you}\:{have}\:{the}\:{right}\:{answer}? \\ $$