What is the number of ways choosing 4 cards from a pack of 52 playing cards?

Number of ways of choosing 4 cards from a pack of 52 cards is the number of combinations of 52 different things taken 4 at time 

∴ Required number of ways = 52C4

= (52 x 51 x 50 x 49)/(4 x 3 x 2 x 1) = 270725

(i) There are four suits, namely diamond, club, spade, heart and each suit has 13 cards. We have to choose 4 cards of the same suit so 4 diamond cards out of 13 diamond cards can be selected in 13C4 ways. Similarly,there are 13C4 ways of choosing 4 spades and 13C4 ways of choosing 4 hearts. 

∴ Required number of ways 

= 13c4 + 13c4 + 13c4 + 13c4

= 4x13C4 = 4 x (13 x 12 x 11 x 10)/(4 x 3 x 2 x 1) = 2860

(ii) There are 13 cards in each suit we have to select 4 cards belonging to 4 different suits. There are 13C1 ways of choosing one card from 13 diamond cards, 13C1 ways of choosing 1 card from 13 cards of spades, 13C1 ways of choosing 1 card from 13 cards of club and 13C1 ways of choosing 1 card from 13 cards of hearts. By multiplication principle, the required number of ways 

= 13C1 x 13C1 x 3C1 x 3C1 

= 13 x 13 x 13 x 13 = 134  =28561

(iii) Face cards means Kings, Queen, Jack there are 4 suits, each suit has 3 face cards. Therefore there are 12 face cards and 4 are to be selected out of 12 cards, in 12C4 ways Required number of ways = 12C4

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.

Solution : Number of ways of choosing 4 cards out of 52
`.^(52)C_(4)=(52xx51xx50xx49)/(4xx3xx2xx1)=270725`.
(i) Choosing 4 cards of the same suit:
We have to draw:
(4 cards from 13 cards of diamond)
or (4 cards from 13 cards of space)
or (4 cards from 13 cards of club)
or (4 cards from 13 cards of heart).
`:.` required number of ways
`(.^(13)C_(4)+.^(13)C_(4)+.^(13)C_(4)+.^(13)C_(4))=4xx.^(13)C_(4)`.
`=(4xx13xx12xx11xx10)/(4xx3xx2xx1)=2860`.
(ii) Choosing 4 cards from 4 different suits:
We have to draw:
(1 card from 13 cards of diamond)
and (1 card from 13 cards of spade)
and (1 card from 13 cards of club)
and (1 card from 13 cards of heart).
`:.` required number of ways
`=(.^(13)C_(1)xx.^(13)C_(1)xx.^(13)C_(1)xx.^(13)C_(1))=(13xx13xx13xx13)=(13)^(4)`.
(iii) Choosing 4 cards out of 12 face cards.
`:.` required number of ways
`=.^(12)C_(4)=(12xx11xx10xx9)/(4xx3xx2xx1)=495`.
(iv) Choosing 2 red cards and 2 black cards:
We have to choose 2 red cards out of 26 and 2 black cards out of 26.
`:.` required number of ways
`=(.^(26)C_(2)x.^(26)C_(2))=((26xx25)/(2))^(2)=(325xx325)=105625`.
(v) Choosing all the 4 cards of the same colour:
We have to choos 4 cards out of 26 black cards or 4 cards out of 26 red cards.
`:.` required number of ways
`=(.^(26)C_(4)+.^(26)C_(4))=(2xx.^(26)C_(4))=(2xx(26xx25xx24xx23)/(4xx3xx2xx1))`
= 29900.

Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these four cards are of the same suit, There are four suits i.e. Diamond, Spade, Heart, Club & 13 cards of each suit Since, they are different cases, So, we add the number of ways Thus, Required number of ways choosing four cards of same suit = 13C4 + 13C4 + 13C4 + 13C4 = 4 × 13C4 = 4 × 13!/(4!(13 − 4)) = 4 × 13!/(4! 9!) = 4 × (13 × 12 × 11 × 10 × 9!)/(4! × 3 × 2 × 1 × 9!) = 4 × (13 × 12 × 11 × 10)/(4 × 3 × 2 × 1) = 2860 ways Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (ii) four cards belong to four different suits, Since, they are the same case, So, we multiply the number of ways Hence, Required number of ways choosing four cards from each suit = 13C1 × 13C1 × 13C1 × 13C1 = (13C1)4 = (13!/1!(13 − 1)!)^4 = (13!/1!12!)^4 = ((13 × 12!)/12!)^4 = (13)4 = 13 × 13 × 13 × 13 = 28561 ways Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (iii) are face cards, King Queen and Jack are face cards Number of face cards in One suit = 3 Total number of face cards = Number of face cards in 4 suits = 4 × 3 = 12 Hence, n = 12 Number of card to be selected = 4 So, r = 4 Required no of ways choosing face cards = 12C4 = 12!/4!(12 − 4)! = 12!/(4! 8!) = (12 × 11 × 10 × 9 × 8!)/(4 × 3 × 2 × 1 × 8!) = (12 × 11 × 10 × 9 )/(4 × 3 × 2 × 1 ) = 495 ways Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (iv) two are red cards and two are black cards, Since, they are the same case, So, we multiply the number of ways Total number of ways choosing 2 red & 2 black cards = 26C2 × 26C2 = (26C2)2 = (26!/(2! (26 − 2)!))^2 = (26!/(2! 24!))^2 = ((26 × 25 × 24!)/(2 × 1 × 24!))^2 = ((26 × 25)/(2 × 1))^2 = (13 × 25)2 = (325)2 = 105625 Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (v) cards are of the same color? Since, choosing red OR black , they are different cases, So, we add the number of ways = 2 × 26!/4!(26 − 4)! = 2 × 26!/(4! 22!) = 2 × (26 × 25 × 24 × 23 × 22!)/(4 × 3 × 2 × 1 × 22!) = 2 × (26 × 25 × 24 × 23)/(4 × 3 × 2 × 1) = 29900

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