Let p be the probability in a pack of playing card two kings are adjacent and q be the probability that no kings are together, then
A. $p = q$
B. $p < q$
C. $p – q = 1$
D. $\dfrac{{48 \times 47 \times 46}}{{52 \times 51 \times 50}}$

Hint: Here, we find the probability by assuming two cards together as one card in the first case, then solve for when no king is together and compare the two values.

Complete step-by-step answer:
Here divide the situations in two cases and solve each of the following cases.
Case - 1: At least two kings are together.
Here if two kings are together then they can be considered as a single card.
Now, total number of card = 51
Number of such permutations = (52 − 1)! = 51!

Case - 2: Assume at least three kings are together.
Here if three kings are together then they can be considered as a single card.
Now, total number of cards = 50
Number of such permutations = (52 − 2)! = 50!
Probability that two kings are adjacent = p
⇒ $p = \dfrac{{51! - 50!}}{{52!}} = \dfrac{{51 - 1}}{{51 \times 52}} = \dfrac{{25}}{{1326}}$
Probability that no two kings are adjacent = q
⇒ $1 - \dfrac{{51!}}{{52!}} = \dfrac{{51}}{{52}}$
⇒ p < q

Hence, option (B) is correct.

Note: In these types of questions, separate the two cases and find the values of both and then compare. Always use the simple concepts for these types of questions and be careful while counting the number of cards. Let us understand the fact used in this solution, why we reduced the number of cards. If two cards are together then both cards are always drawn together or the two cards cannot be drawn separately. So, in this way the favourable outcome of the king is only three as separately only 3 kings can be drawn. Hence reduction of 1 card is there in total number of cards.