Here's how I would do that: there are 52 cards, 4 aces, four kings, and a jack. The probability the first card drawn is an ace is 4/52= 1/13. There are then 51 cards left, flur of which are kings. The probability that the second card you draw is a king is 4/51. There are then 50 cards left, 4 of which are jacks. The probability the third card you draw is a jack is 4/50= 2/25. The probability of drawing "ace, king, jack"in that orderis (1/13)(4/51)(2/25)= 8/(13*51*25). But if you look at "jack, ace, king" or any other specific order you will see that while you have different fractions, you have the same numerators and the same denominators in different orders so the same probability. There are 3!= 6 such orders so there the probability of drawing an ace, king, and jack is 6(8/(13*51*25)).

The reason for " " is that a total of 12 "aces, kings, and jacks" and you are drawing one of them. The reason for the " " is that whether that is an ace, king, or jack, there are then 8 of the remaining kind of card you are looking for (if the first card drawn was a jack, there are 8 aces and kings left) and you want to draw one of them. The reason for the " " is that whichever of ace, jack, or king, the first two cards are, there are 4 cards of the remaining type and you want to draw 1 of them.Solution given:- There are 4 aces, 4 king and 4 jacks and their selection can be made in the following ways:

^{12}C_{1}X^{8}C_{1}X^{4}C_{1}= 12 X 8 X 4.

Total selections can be made =^{52}C_{3}= 52 X 51 X 50.

Therefore required probability =

I don't understand why are we taking^{12}C_{1}X^{8}C_{1}X^{4}C_{1}= 12 X 8 X 4 instead of^{4}C_{1}X^{4}C_{1}X^{4}C_{1}= 4 x 4 X 4 for the numerator. Since, we are selecting 1 ace from 4 aces, 1 king from 4 kings and 1 jack from 4 jacks shouldn't we be taking^{4}C_{1}X^{4}C_{1}X^{4}C_{1}= 4 x 4 X 4 for the favourable events ? Please advice on the above.

Thanks in advance !