Question

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

Answer 1

Hint: We need to find the number of ways the cards in a deck can be arranged in such a way that, all the cards of the same suit are together. This means that we have to arrange 13 cards in a suit in 13! ways. There are 4 such suits and the 4 suites can be arranged in 4! ways. The product of these arrangements gives the total number of arrangements.

Complete step by step Answer:

We have a deck of 52 cards with 4 suits of 13 cards each.
We can consider 4 suits as 4 different packs with 13 cards each. Then four suits then can be arranged in \[{\text{4!}}\] ways. We can consider a pack of 13 cards. These 13 cards are not identical and we can arrange the internally in 13! ways. We have four such packs of 13 cards and each of them can be arranged by 13! ways.
Therefore, the number of ways the four suits can be arranged internally is \[13! \times 13! \times 13! \times 13! = {\left( {13!} \right)^4}\]
Therefore, we get the total number of ways of arranging a deck of cards such that all the cards of the same suit are together by taking the product of the number of ways of arranging the four suits internally and the number of ways 4 packs of 13 cards can be arranged.
Therefore, the required number of ways is given by the product ${\left( {{\text{13!}}} \right)^{\text{4}}}{\text{.4!}}$
So, the correct answer is option C.

Note: We must consider all the possible arrangements, that is we must consider the cases of internal arrangements within a suit as well as the different arrangements of all the suits. We must not take the cards of the same suit as identical cards. All the cards in a deck is different in one way or another so we must take its permutations. We must take the product to find the number of ways of arranging the cards of 4 suits internally and not the sum. We need to find the exact value of the answer as it is time-consuming and increases the chances of making mistakes as there is a lot of multiplication involved. In the options also the number of ways is given in terms of the product of factorial.