To solve this problem, we need to consider combinations, as we are drawing cards without replacement and the order does not matter.

Step 1: Analyze the Problem

We have a deck of 117 cards with 9 unique suits, each containing 13 cards (Ace, 2 through 10, Jack, Queen, King).

• Number of Aces in the deck: Since there are 9 suits, each having 1 Ace, we have a total of 9 Aces in the deck.
• Number of "Hearts" in the deck: The "Hearts" suit consists of 13 cards, including one Ace of Hearts.

Step 2: Define the Required Conditions

We want to draw 4 cards without replacement such that:

1. Exactly 3 of the cards are Aces.
2. Exactly 1 of the cards is a Heart.

Step 3: Calculate the Combinations

1. Choose 3 Aces from the 9 available Aces:
   The number of ways to choose 3 Aces out of 9 is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items, and $k$ is the number of items to choose.

   $C(9, 3) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

2. Choose 1 Heart from the remaining Hearts: After choosing 3 Aces, we need to ensure that exactly one card is a Heart, but it should not overlap with the Aces we selected.

   • If we selected the Ace of Hearts as one of the 3 Aces: This means we have 12 remaining Heart cards to choose from.

     $C(12, 1) = 12$

   • If we did not select the Ace of Hearts as one of the 3 Aces: This means we have all 13 Heart cards available to choose from.

     $C(13, 1) = 13$

Step 4: Calculate Total Combinations

We need to consider both cases:

1. Ace of Hearts is included in the 3 Aces: [ \text{Number of ways} = C(8, 2) \times C(12, 1)

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