Probability of certain cards in 10 card poker

Question

Find the probability of being dealt seven clubs and three cards with one card of each other remaining suit in 10​-card poker.

Answer 1

Given: 10 card hand, no discarding.

Assumption: 1 deck

Reminders for the non-gamblers among us: 13 cards each in each of 4 suits, (2, 3, . . . J, K, Q, A)

1.) Consider the denominator. How different combinations can you get from a 10 card hand out of a 52 card deck. ₅₂C₁₀

2,) Consider the numerator. What outcomes are asked for and how are they related to each other.

1. N ways that 7 of 10 cards will be clubs. ₁₃C₇
2. N Ways that each remaining card will be from a different suit among the 3 suits remaining.
3. N ways to draw one diamond: ₁₃C₁
4. N ways to draw one heart: ₁₃C₁
5. N ways to draw one spade: ₁₃C₁
6. How to these outcomes related to each other.
7. AND, so we'll want multiplication
8. So the numerator is: ₁₃C₇ * 3 * ₁₃C₁
9. The requested probability is then:

(₁₃C₇ * 3 * ₁₃C₁)/₅₂C₁₀