From the uploaded image, the problem involves counting the number of ways a run of three cards in order of the same suit can be drawn from a standard 52-card deck of cards. Let’s break it down step by step.

Explanation:

1. Part (a): How many ways can the suit be chosen?

   • There are 4 suits in a standard deck (hearts, diamonds, clubs, spades).
   • You can choose one of these suits in $4$ ways.

2. Part (b): How many ways can the three cards from the suit be chosen?

   • A "run of three cards" refers to 3 consecutive cards in a sequence (e.g., 2, 3, 4 of the same suit).
   • The sequences start with any card from Ace (low card) to King (high card), but only up to 12 possible starting cards. This is because a run starting at King (13) would require 14, which doesn't exist.
   • Therefore, there are $12$ possible runs in each suit.

3. Part (c): How many ways can the run of three be chosen?

   • Since you can choose a suit in $4$ ways, and there are $12$ runs per suit, the total number of runs is: $4 \times 12 = 48$
   • There are $48$ ways to choose the run of three cards in total.

Let me know if you'd like further clarification! Here's more to consider:

Questions to expand understanding:

1. What if the cards do not have to be in the same suit? How does the calculation change?
2. Can a run of cards include an Ace as both high (e.g., Q, K, A) and low (e.g., A, 2, 3)?
3. What is the probability of being dealt a run of three cards in order in a random hand of 3 cards?
4. How does this problem change if the deck contains jokers or extra cards?
5. What is the mathematical difference between permutations and combinations in this context?

Tip:

Always carefully define terms in probability or combinatorics problems (e.g., "run," "suit") to ensure correct interpretation.