Question

The number of ways in which 52 playing cards can be divided into 4 sets, three of them having 17 cards each and fourth is having just one card is:
A. \[\dfrac{{52!}}{{{{(17!)}^3}}}\]
B. \[\dfrac{{52!}}{{{{(17!)}^3}3!}}\]
C. \[\dfrac{{51!}}{{{{(17!)}^3}}}\]
D. \[\dfrac{{51!}}{{{{(17!)}^3}3!}}\]

Answer 1

Hint: First give one card to one player and then start distributing the cards among others one by one to find the probability. So we will have one player with one card, and we will have 3 players with 17 cards here, we see the number of ways of doing this.

Complete step by step solution: We want to distribute a pack of 52 cards among four players such that three get 17 cards each and one gets 1 card.
Let us first give 1 card to the fourth player. This can be done in 52 ways.
Let us now give 17 out of the remaining 51 cards to the first player. This can be done in \[{}^{51}{C_{17}}\] ways.
Let us now give 17 out of the remaining 34 cards to the second player. This can be done in \[{}^{34}{C_{17}}\] ways.
The remaining 17 cards can be given to the third player only 1 way.
 \[ \Rightarrow \] The total numbers of ways of distributing the cards is,
 \[52 \times {}^{51}{C_{17}} \times {}^{34}{C_{17}} \times 1\]
On using the formula \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\] we get,
 \[ = 52 \times \dfrac{{51!}}{{17!34!}} \times \dfrac{{34!}}{{17!17!}}\]
On further simplification we get,
 \[ = \dfrac{{52!}}{{{{(17!)}^3}}}\]

Hence option (A) is correct.

Note: The thing we should keep in mind is that the persons we are referring to are all random. Like the 1st person who is having 1 card can also have 17 cards. Similarly, the other players who are having 17 cards can also have 1 card too.