Question

The number of ways in which a pack of 52 cards of four different suits is distributed equally among 4 players so that each may have ace, king, queen, and knave of the same suit is:
A. $ \dfrac{{4!(36!)}}{{{{(9!)}^4}}} $
B. $ \dfrac{{(36!)}}{{{{(9!)}^4}}} $
C. $ \dfrac{{2(36!)}}{{{{(9!)}^4}}} $
D. $ \dfrac{{3(36!)}}{{{{(9!)}^4}}} $

Answer 1

Hint: The cards are distributed equally among players with each having face cards and knaves of the same suit and this distribution can be in any order, so we will use factorial.
Then when these cards are distributed,we will be applying combinations on the remaining cards to be distributed among players. Each deck has 4 suits namely diamonds, clubs, hearts, and spades.

Complete step-by-step answer:
Face cards + Knave = 4
Distributed among 4 players.
As these can be any order, these can be distributed as 4! ……… (1)
Now,
Cards distributed = \[\left( {4 \times 4} \right)\] [4 face cards among 4 players]
             $ = 16 $
Remaining cards \[ = {\text{ }}52 - 16\](Total cards in deck = 52)
             $ = 36 $
We have to divide 36 cards among these players, for this, we will use combinations:
The number of cards has to be distributed equally among 4 players, which gives
 $ \dfrac{{52}}{4} = 13 $
Each player is supposed to get 13 cards out of which 4 face cards are received, number of cards remaining.
\[13 - 4 = 9\]
Therefore, the value of r in combination will be 9.
Applying combination $ \left( {{}^n{C_r}} \right) $ where n is number of cards and r is the card each player faces.
\[{}^{36}{C_9} \times {}^{36 - 9}{C_9} \times {}^{36 - 18}{C_9} \times {}^{36 - 27}{C_9}\]
Applying the formula to solve the combination.

\[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], we get:
 $ \dfrac{{36!}}{{9!27!}} \times \dfrac{{27!}}{{9!8!}} \times \dfrac{{18!}}{{9!9!}} \times \dfrac{{9!}}{{9!0!}} $
\[ = \dfrac{{36!}}{{{{\left( {9!} \right)}^4}}}.....(2)\]

Therefore, the total number of ways in which a pack of 52 cards can be distributed equally among 4 players each from (1) and (2) is\[\dfrac{{4!\left( {36!} \right)}}{{{{\left( {9!} \right)}^4}}}\], option A.
So, the correct answer is “Option A”.

Note: In a deck of 52 cards:
There are 4 suits namely hearts, clubs, diamonds, and spades.
Each suit has 13 cards.
There are 3 face cards in each suit, these total face cards: $ 3 \times 4 = 12 $
Out of 52 half cards are black (spade & club) and half are red (Diamond & heart)