Question

What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
$\left(i\right)$ four cards are of the same suit.
$\left(ii\right)$ four cards belong to four different suits.
$\left(iii\right)$ are face cards.
$\left(iv\right)$ two are red cards and two are black cards.
$\left(v\right)$ cards are of the same colour?

Answer 1

Finding number of ways choosing $4$ cards from $52$ cards.
$n=52$ and $r=4$
$\therefore$ Required number of way $={}^{52}{C}_4$
$=\frac{52!}{4!(52-4)!}$
$=\frac{52\times 51\times 50\times 49\times 48!}{4!48!}$
$=\frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}$
$=270725$

$\left(i\right)$ Finding number of choosing $4$ cards from the same suit
$\therefore$ There are four suits, Diamond, Spade, Heart and Club and each suit has $13$ cards.
$\therefore$ Number of ways of choosing $4$ diamonds $={}^{13}{C}_4$
Number of ways of choosing $4$ spades $={}^{13}{C}_4$
Number of ways of choosing $4$ hearts $={}^{13}{C}_4$
Number of ways of choosing $4$ clubs $={}^{13}{C}_4$

$\therefore$ Required number of ways of choosing four cards of same suit $=4{\times }^{13}{C}_4$
$=4\times \frac{13!}{4!(13-4)!}$
$=4\times \frac{13\times 12\times 11\times 10\times 9!}{4\times 3\times 2\times 1\times 9!}$
$=2860\text{ways}$.

$\left(ii\right)$ Finding number of choosing 4 cards belong to four different suits
Number of cards is each suit $=13$
Number of ways for choosing $1$ card of diamond $={}^{13}{C}_1$
Number of ways for choosing $1$ card of spades $={}^{13}{C}_1$
Number of ways for choosing $1$ card of clubs $={}^{13}{C}_1$
Number of ways for choosing $1$ card of clubs $={}^{13}{C}_1$
$\therefore$ Required number of ways of choosing four cards of same suit $={}^{13}{C}_1\times {}^{13}{C}_1\times {}^{13}{C}_1\times {}^{13}{C}_1$.
$=({}^{13}{C}_1{)}^4$
$={\left(\frac{13!}{(13-1)!1!}\right)}^4$
$={\left(\frac{13\times 12!}{12!1!}\right)}^4$
$={\left(\frac{13}{1}\right)}^4$
$=(13{)}^4$
$=28561\text{ways}$.

$\left(iii\right)$ Finding number of ways of choosing face cards.
Total number of face cards $\left(n\right)=12$
Number of cards to be selected $=r=4$
$\therefore$ Required number of ways choosing face cards $={}^{12}{C}_4$

$=\frac{12!}{4!\times (12-4)!}$
$=\left(\frac{12!}{4!\times 8!}\right)$
$=\left(\frac{12\times 11\times 10\times 9\times 8!}{4\times 3\times 2\times 1\times 8!}\right)$
$=11\times 5\times 9$
$=495\text{ways}$

$\left(iv\right)$ Finding number of ways choosing $4$ cards of which two are red and two are black cards.
$\therefore$ Number of Red cards $=26$
Number of Black cards $=26$
$\therefore$ Number of ways of choosing $2$ red and $2$ black cards
$={}^{26}{C}_2\times {}^{26}{C}_2$
$=\frac{26!}{2!\times 24!}\times \frac{26!}{2!\times 24!}$
$=\frac{26\times 25\times 24!}{2\times 1\times 24!}\times \frac{26\times 25\times 24!}{2\times 1\times 24!}$
$=325\times 325$
$=105625\text{ways}$.

$\left(v\right)$ Finding number of ways of choosing 4 cards are of the same colour
Number of ways of choosing $4$ red card $={}^{26}{C}_4$
Number of ways of choosing $4$ black card $={}^{26}{C}_4$
$\therefore$ Required number of ways of choosing $4$ red card or $4$ Black card
$={}^{26}{C}_4+{}^{26}{C}_4$
$=2\times {}^{26}{C}_4$
$=2\times \frac{26\times 25\times 24\times 23\times 22!}{4!(26-4)!}$
$=\frac{2\times 26\times 25\times 24\times 23\times 22!}{4\times 3\times 2\times 1\times 22!}$
$=2\times 26\times 25\times 23$
$=29900\text{ways}$.