Question

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?

Answer 1

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;

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

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