The number of ways of dividing $52$ cards amongst four players so that three players have $17$ cards each and the fourth player has just one card,is

  • A
    $\frac{52!}{(17!)^3}$
  • B
    $52!$
  • C
    $\frac{52!}{17!}$
  • D
    None of these

Explore More

Similar Questions

The number of ways of dividing $15$ persons into $3$ groups containing $3, 5$ and $7$ persons such that two particular persons are not included in the $5$-person group is:

There are $8$ different coloured balls and $8$ bags having the same colours as that of the balls. If one ball is placed at random in each one of the bags,then the probability that $5$ of the balls are placed in the respective coloured bags,is

$n$ distinct items $1, 2, 3, \dots, n$ are arranged in $n$ distinct positions $1, 2, 3, \dots, n$. What is the probability that at least three items are in their correct positions?

Difficult
View Solution

Six cards and six envelopes are numbered $1, 2, 3, 4, 5, 6$. Cards are to be placed in envelopes such that each envelope contains exactly one card,no card is placed in the envelope bearing the same number,and the card numbered $1$ is always placed in the envelope numbered $2$. The number of ways this can be done is:

The set $S = \{1, 2, 3, \dots, 12\}$ is partitioned into three sets $A, B, C$ of equal size such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = C \cap A = \phi$. In how many ways can $S$ be partitioned?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo