From a pack of 52 playing cards, three are drawn at random. Find the probability of drawing a king, a queen and jack.

Question

A simple question but the solution is confusing me. The answer I obtained was $$p = 3! \times 4/52 \times 4/51 \times 4/50$$

The first 3! is for the order of king, queen, jack. $4/52$ is the probability of drawing a king, $4/51$ is the probability of drawing a queen after a king is drawn and $4/50$ is the probability of drawing a jack once both king and queen are drawn. But the book takes the solution as

$$p = 3! \times 4/52 \times 3/51 \times 2/50$$

Can anyone please explain how this comes?

Thanks in advance.

Answer 1

The probability of drawing a king, queen, and jack is;

$$\frac{3! 4^3}{52\,51\,50}$$

The probability of drawing a king, queen, and jack of different suits is:

$$\frac{3!~4!}{52\,51\,50}$$

Answer 2

If you don't care about the order, answer is $$\left(\frac{12}{52}\right)\times \left(\frac{6}{51}\right)\times \left(\frac{2}{50}\right)$$

Because you have $12$ ways to pick the first card (K,Q, or J of any suit), then $6$ ways to pick the second card ( two remaining unpicked faces of 3 remaining suits), then finally $2$ ways to pick the last card (the remaining unpicked face with $2$ remaining suits). The denominator decrements each time in accordance with assumption of no replacement of previously selected cards.

This answer is equivalent to what was previously stated as the book answer: $$p=3! \times \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50}$$