Answer to Question #206530 in Civil and Environmental Engineering for Joshua

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?


1
Expert's answer
2021-06-14T17:04:01-0400

Part A;

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

40C8=40!8!(108)!^{40}C_8= \frac{40!}{8!(10-8)!}

=403938............321(876.......321(21)=\frac{40*39*38............*3*2*1}{(8*7*6*.......*3*2*1(2*1)}

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

Part B;

8C18C28C38C18C1=8!1!(81)!8!2!(82)!8!3!(83)!8!1!(81)!8!1!(81)!^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)!}

8!7!8!26!8!65!8!7!8!7!=8285688\frac{8!}{7!}* \frac{8!}{2*6!}* \frac{8!}{6*5!}* \frac{8!}{7!}* \frac{8!}{7!}=8*28*56*8*8

=802816=802816

Part C;

32C8=32!8!(328)!^{32}C_8= \frac{32!}{8!(32-8)!}

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


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