The number of ways of dividing 52 cards among | vedclass

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

Question

(A) To divide $52$ cards equally among four players,each player receives $13$ cards.The number of ways to choose $13$ cards for the first player from

Vedclass · Accepted Answer

$\frac{52!}{(13!)^4}$

Vedclass · Answer

(A) To divide $52$ cards equally among four players,each player receives $13$ cards.The number of ways to choose $13$ cards for the first player from $52$ is $^{52}C_{13}$.The number of ways to choose $13$ cards for the second player from the remaining $39$ is $^{39}C_{13}$.The number of ways to choose $13$ cards for the third player from the remaining $26$ is $^{26}C_{13}$.The number of ways to choose $13$ cards for the fourth player from the remaining $13$ is $^{13}C_{13}$.Total number of ways = $^{52}C_{13} 	imes ^{39}C_{13} 	imes ^{26}C_{13} 	imes ^{13}C_{13}$$= \frac{52!}{39! \cdot 13!} 	imes \frac{39!}{26! \cdot 13!} 	imes \frac{26!}{13! \cdot 13!} 	imes \frac{13!}{0! \cdot 13!}$$= \frac{52!}{(13!)^4}$.

The number of ways of dividing $52$ cards among four players equally is:

Similar Questions

The number of ways in which $9$ persons can be divided into three equal groups is

On the occasion of the Deepawali festival,each student of a class sends greeting cards to all the others. If there are $20$ students in the class,then the total number of greeting cards exchanged by the students is:

The number of ways in which ten candidates $A_1, A_2, ....... A_{10}$ can be ranked such that $A_1$ is always above $A_{10}$ is

There are $12$ members in a group. In how many ways can a team of $6$ members be selected such that a particular member is always included?

How many numbers can be formed with the digits $1, 2, 3, 4, 3, 2, 1$ so that odd digits always occupy the odd places?

Vedclass Test Series

Exam Paper Generator

Online Exam Module