Probability of ordered sequence of cards

Question

When selecting three cards (without replacement) in an ordered sequence, what is the probability that the rank of the first card is strictly smaller than the rank of the second card which, in turn, is strictly smaller than the rank of the third card?

I don’t know how to approach this problem. Any hints?

Answer 1

Choose 3 distinct card ranks from 13. Choose one card from each rank. Place them in ascending order (there is only one order for them that works). Divide by the total number of ways to choose 3 cards in order.

Answer: $$\dfrac{176}{1275}$$

Answer 2

If you have any ties, you fail. If there are no ties, one order in six makes you a winner.

Answer 3

When drawing without replacement first 3 different rank cards and there are 4 possible cards of each rank. Number of ways will be = (13!/10!.3! )(4^3)

Whereas sample space will be all possible ways=52*51*50 Hence probability is 176/1275

Answer 4

P(Choosing 3 distinct card ranks from 13) : 52/52* 48/51 * 44/50

P(Ways you can arrange in descending order) : 1/6 * 48/51 * 44/50 = 176/1275

thanks for the hints!