Question

What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many ways these

(i) four cards are of the same suit?

(ii) four cards are belongs to four different suits?

(iii) four cards are of the same colour?

Answer 1

Four cards can be chosen from 52 cards in ${}^{52}{C}_4$ ways, i.e. $\frac{52!}{48!4!}=\frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}=270725$ ways (i) There are four suits (diamond, spade, club and heart) of 13 cards each. Therefore, there are ${}^{13}{C}_4$ ways of choosing 4 diamonds cards. ${}^{13}{C}_4$ ways of choosing 4 club cards, ${}^{13}{C}_4$ways of choosing 4 spade cards and ${}^{13}{C}_4$ ways of choosing 4 heart cards.

$\therefore$ Required number of ways

= ${}^{13}{C}_4{+}^{13}{C}_4{+}^{13}{C}_4{+}^{13}{C}_4$

= $4{\times }^{13}{C}_4=4\times \frac{13!}{4!9!}$

= $\frac{4\times 13\times 12\times 11\times 10}{4\times 3\times 2\times 1}=2860$

(ii) There are 13 cards in each suit. Four cards drawn belong to four different suits means one card is drawn from each suit. Out of 13 diamond cards, one card can be drawn in ${}^{13}{C}_1$ ways. Similarly, there are ${}^{13}{C}_1$ ways of choosing one club card, ${}^{13}{C}_1$ way sof choosing one spade card and ${}^{13}{C}_1$ ways of choosing one heart card.

$\therefore$ Number of ways of selecting one card from each suit = ${}^{13}{C}_1{\times }^{13}{C}_1{\times }^{13}{C}_1{\times }^{13}{C}_1=(13{)}^4=28561$

= $4{\times }^{13}{C}_4=4\times \frac{13!}{4!9!}$

= $\frac{4\times 13\times 12\times 11\times 10}{4\times 3\times 2\times 1}=2860$

(iii) Out of 26 red cards, 4 cards can be selected in ${}^{26}{C}_4$ ways. Similarly, 4 black cards can be selected in ${}^{26}{C}_4$ ways.

Hence, 4 red or 4 black cards can be selected in ${}^{26}{C}_4{+}^{26}{C}_4$ ways

= $2\times {}^{26}{C}_4$=$\frac{2\times 26!}{4!22!}$

= $\frac{2\times 26\times 25\times 24\times 23}{4\times 3\times 2\times 1}$

= 29900 ways