What is the number of ways of choosing 4 card | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The total number of ways to choose $4$ cards from $52$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.Total ways $= ^{52}C_

Vedclass · Accepted Answer

$270725$ and $105625$

Vedclass · Answer

(A) The total number of ways to choose $4$ cards from $52$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.Total ways $= ^{52}C_{4} = \frac{52 	imes 51 	imes 50 	imes 49}{4 	imes 3 	imes 2 	imes 1} = 270725$.There are $26$ red cards and $26$ black cards in a deck.To choose $2$ red cards from $26$ and $2$ black cards from $26$,the number of ways is $^{26}C_{2} 	imes ^{26}C_{2}$.$^{26}C_{2} = \frac{26 	imes 25}{2 	imes 1} = 325$.So,the required number of ways $= 325 	imes 325 = 105625$.

What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these are two red cards and two black cards?

Similar Questions

$A$ student is asked to answer $10$ out of $13$ questions in an examination such that he must answer at least four questions from the first five questions. Then the total number of possible choices available to him is

$A$ student is asked to answer $10$ out of $13$ questions in an examination such that he must answer at least four questions from the first five questions. The number of choices available to him is

In how many ways can we select any $4$ letters from the word $CORGOO$?

The number of non-negative integral solutions of $x_1+x_2+x_3+x_4=10$ is

In how many ways can $6$ '$+$' signs and $4$ '$$' signs be arranged in a row such that no two '$$' signs occur together?

Vedclass Test Series

Exam Paper Generator

Online Exam Module