Question

The number of arrangements of all the cards in a complete deck, such that the cards of the same suit are together.
A. $\dfrac{{52!}}{{4!}}$
B. $\dfrac{{{{\left( {13!} \right)}^4}}}{{4!}}$
C. ${\left( {13} \right)^4}4!$
D. $\dfrac{{52!}}{{{{\left( {13!} \right)}^4}}}$

Answer 1

We have to determine the number of ways such that the cards can be arranged in a way such that all the same suits are together. There are 13 cards of suit in a deck. Each card can be arranged in 13! ways. Also, multiply this with the number of ways 4 suits can be arranged, that is, 4!.

Complete step by step solution:
There are 52 cards in a deck and in a deck there are 13 cards of each suit.
There are a total 4 suits in the pack of 52 cards.
Consider each suit as a different pack with 13 cards each.
If there are $n$ objects and we have to arrange in $n$ places, then the number of ways doing it is $n!$ ways.
Now, 13 different cards have to be arranged in 13 different places.
Therefore, the number of ways in which 13 cards can be arranged is 13!
But, there are 4 suits then the total number of ways $13! \times 13! \times 13! \times 13! = {\left( {13!} \right)^4}$
Therefore, now each suit can also arrange among themselves.
Then, ${\left( {13!} \right)^4}4!$

Hence, option C is correct.

Note:
We must know that there are 52 cards in a deck. There are 4 suits in a deck in which there are 13 cards in each suit. Also, all cards are different in a suit, that is there are no identical cards in the same deck. Hence, we will use the condition that if there are $n$ objects and we have to arrange in $n$ places, then the number of ways doing it is $n!$ ways.