The face cards are removed from a pack of 52 cards then 4 cards are drawn one by one from the remaining 40 cards

Question

The face cards are removed from a pack of $52$ cards then $4$ cards are drawn one by one from the remaining $40$ cards what is the probability that four cards belong to different suits and different denomination.

The answer given is = 21/9139

My solution is as shown...

For a particular case the 1st card be diamond, 2nd be spade, 3rd be club and 4th be heart.

For this particular case the probability is given as (using conditional probability or general multiplication rule)

= (10/40).(9/39).(8/38).(7/37)

=21/9139

Now 4! such cases are possible which are mutually exclusive where the order of diamond ,club, heart and spade can be altered, so the required probability is

= 4! (21/9139)

=504/9139

Please say whether I am right or wrong.

Rectify me if I am.

Answer 1

You are correct. Let's confirm your answer.

It does not matter which card we draw first.

Once it has been selected, there are $39$ cards remaining. Of these, nine have the same suit as the first card and three other cards have the same rank (denomination) as the first card. Hence, there are $39 - 12 = 27$ favorable selections for the second card.

Once two cards with different ranks and different suits have been selected, $38$ cards remain. Of these, $18$ belong to one of the suits that has already been selected. Of the $20$ cards in the remaining two suits, $4$ cards share a rank with one of the selected cards. Hence, there are $38 - 18 - 4 = 16$ favorable selections for the third card.

Once three cards with different ranks and different suits have been selected, $37$ cards remain. Of these, $27$ cards share a suit with one of the selected cards. Of the $10$ cards in the remaining suit, $3$ cards share a rank with one of the selected cards. Hence, there are $37 - 27 - 3 = 7$ favorable selections for the fourth card.

Thus, the probability that four cards with different ranks and different suits are selected is $$1 \cdot \frac{27}{39} \cdot \frac{16}{38} \cdot \frac{7}{37} = \frac{504}{9139}$$ as you found.