Question #206530

a card game uses 40 unique cards with 5 suits (diamonds, hearts, clubs, spades and thunder. Each suit is numbered from 1 to 8. to play the game, a player must hold 8 cards which may be sorted anyway the player choose. a) How many 8-card hands are possible? b) How many 8-card hands consisting of 1 diamonds, 2 hearts, 3 clubs, 1 spades and 1 thunder are possible? c) How many 8-card hands consisting of no thunder are possible?

Expert's answer

Part A;

"^n C_r= \\frac{n!}{r!(n-r!)}"

"^{40}C_8= \\frac{40!}{8!(10-8)!}"

"=\\frac{40*39*38............*3*2*1}{(8*7*6*.......*3*2*1(2*1)}"

"=76904685" possibilities of 8-card hands are possible

Part B;

"^8C_1*^8C_2*^8C_3*^8C_1*^8C_1=\\frac{8!}{1!(8-1)!}* \\frac{8!}{2!(8-2)!}* \\frac{8!}{3!(8-3)!}*\\frac{8!}{1!(8-1)!}*\\frac{8!}{1!(8-1)!}"

"\\frac{8!}{7!}* \\frac{8!}{2*6!}* \\frac{8!}{6*5!}* \\frac{8!}{7!}* \\frac{8!}{7!}=8*28*56*8*8"

"=802816"

Part C;

"^{32}C_8= \\frac{32!}{8!(32-8)!}"

"=\\frac{32!}{8!24!}"

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