Question

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

Answer 1

The number of cards in a pack of cards is 52 which consists of $4$ suits of $13$ cards. We solve this problem using some simple method of probability. The probability of an event is defined as $P(E)= \dfrac{\text{Number of favourable outcomes of E}}{\text{Total number of possible outcomes of E}}$. We can also solve these types of problems using permutation and combination.

Complete step-by-step solution:
First we calculate the total number of ways we can choose $4$ cards out of 52 cards. For this purpose, we calculate ${}^{52}{C_4} = \dfrac{{52!}}{{4!\left( {52 - 4} \right)!}} = \dfrac{{52!}}{{4!\left( {48} \right)!}}$
i.e. $^{52}{C_4} = \dfrac{{52 \times 51 \times 50 \times 49 \times 48!}}{{24 \times 48!}}$
i.e. $^{52}{C_4} = 270725$
PART(i): We choose four cards of the same suit.
We have four suits in a pack of cards, namely Diamond, Club, Heart and Spade. Each suit contains $13$ cards each. Now let us choose $4$cards out of the $13$ diamond cards in $^{13}{C_4}$ ways. In the very same process, we choose $4$cards out of the $13$ club cards in $^{13}{C_4}$ ways. Same as $^{13}{C_4}$ ways for heart and $^{13}{C_4}$ ways for spade.
Therefore the total number of ways of choosing$4$ cards of the same suit is given by ${ = ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4} = 4{ \times ^{13}}{C_4}$
i.e. $4 \times \dfrac{{13!}}{{4!\left( {13 - 4} \right)!}}$
$= 4 \times \dfrac{{13 \times 12 \times 11 \times 10 \times 9!}}{{24 \times 9!}} = 2860$
Hence the total number of ways of choosing four cards of the same suit is $2860$ .
PART(ii): We choose four cards belonging to four different suits.
In this case we choose a $1$ card each from every suit. We can choose one card from a suit in $^{13}{C_1}$ ways and for four different suits, the total number of ways we obtain is $^{13}{C_1}{ \times ^{13}}{C_1}{ \times ^{13}}{C_1}{ \times ^{13}}{C_1} = {\left( {^{13}{C_1}} \right)^4} = {\left[ {\dfrac{{13!}}{{1!\left( {13 - 1} \right)!}}} \right]^4} = {\left[ {\dfrac{{13 \times 12!}}{{1 \times \left( {12} \right)!}}} \right]^4} = {\left( {13} \right)^4}$
Hence the total number of ways of choosing four cards from four different suits is ${\left( {13} \right)^4}$ .
PART(iii): We choose four face cards.
We know there are twelve face cards, four kings, four queens and four jokers. Amongst these $12$ face cards we choose $4$ cards in $^{12}{C_4}$ ways i.e. $\dfrac{{12!}}{{4!\left( {12 - 4} \right)!}} = \dfrac{{12 \times 11 \times 10 \times 9 \times 8!}}{{4!\left( 8 \right)!}} = 495$ ways.
PART(iv): We choose four cards of the same colour.
We know out of the $52$ cards, there are $26$ black cards and $26$ red cards. So we either have $4$ black cards or $4$ red cards.
Either we choose $4$ out of $26$ black cards in $^{26}{C_4}$ ways or, we choose $4$ out of $26$ red cards in $^{26}{C_4}$ ways.
Therefore the required number of selection is $^{26}{C_4}{ + ^{26}}{C_4} = 2{ \times ^{26}}{C_4} = 2 \times \dfrac{{26!}}{{4!\left( {26 - 4} \right)!}} = 2 \times \dfrac{{26 \times 25 \times 24 \times 23 \times 22!}}{{24 \times \left( {22} \right)!}} = 2 \times 14950 = 29900$ ways.
PART(v): We choose two red cards and two black cards.
So we choose $2$ cards out of the $26$ red cards and also we have to choose $2$ cards out of the $26$ black cards.
We choose $2$ out of $26$ red cards in $^{26}{C_2}$ ways and $2$ out of $26$ black cards in $^{26}{C_2}$ ways.
So the total number of selections is given by $^{26}{C_2}{ + ^{26}}{C_2} = 2{ \times ^{26}}{C_2} = 2 \times \dfrac{{26!}}{{2!\left( {26 - 2} \right)!}} = 2 \times \dfrac{{26 \times 25 \times 24!}}{{2 \times \left( {24} \right)!}} = 650$ways.

Note: Note that we used the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so often in this problem, where $^n{C_r}$is defined as the number of combinations obtained when choosing r things out of a total number of n things. And the term $r!$ is called “r factorial” and is defined by $r! = r \times (r - 1) \times (r - 2) \times ......2 \times 1$ where r is a positive integer>$1$ . Also we should know that Combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter.