To solve the problem of determining how many different 5-card hands can be formed from a deck that contains all four suits (Clubs ♣, Spades ♠, Hearts ♥, and Diamonds ♦) but only the cards with values 5, 6, 7, 8, 9, and 10, with the condition that each hand must contain at least one Spade and at least two 10s, we can break down the problem into steps.

Step 1: Understand the Deck Composition

• The deck consists of cards with values 5, 6, 7, 8, 9, and 10.
• Each value has 4 cards (one for each suit: ♣, ♠, ♥, ♦).
• There are a total of $6 \times 4 = 24$ cards in the deck.

Step 2: Choose the 10s

To satisfy the condition that the hand contains at least two 10s:

• Case 1: The hand contains exactly two 10s.
• Case 2: The hand contains exactly three 10s.
• Case 3: The hand contains exactly four 10s.

Case 1: Two 10s

• Number of ways to choose 2 suits for the 10s out of 4 suits: $\binom{4}{2} = 6$.

Case 2: Three 10s

• Number of ways to choose 3 suits for the 10s out of 4 suits: $\binom{4}{3} = 4$.

Case 3: Four 10s

• There is only one way to choose all 4 suits for the 10s: $\binom{4}{4} = 1$.

Step 3: Ensure At Least One Spade in the Hand

For each case, we must ensure that the remaining cards (not the 10s) include at least one spade.

Case 1: Two 10s

• After choosing 2 tens, 3 more cards must be selected from the remaining 20 cards (excluding the 10s), but we need at least one spade.
• Number of ways to choose these 3 cards, ensuring at least one spade, can be calculated using the complementary method:
  • Total ways to choose 3 cards from 20: $\binom{20}{3}$.
  • Subtract the number of ways to choose 3 cards without any spades (there are 15 non-spades): $\binom{15}{3}$.
  • Therefore, the number of valid combinations: $\binom{20}{3} - \binom{15}{3}$.

Case 2: Three 10s

• After choosing 3 tens, 2 more cards must be selected from the remaining 20 cards, ensuring at least one spade.
• Using the complementary method again:
  • Total ways to choose 2 cards from 20: $\binom{20}{2}$.
  • Subtract the number of ways to choose 2 cards without any spades (15 non-spades): $\binom{15}{2}$.
  • Therefore, the number of valid combinations: $\binom{20}{2} - \binom{15}{2}$.

Case 3: Four 10s

• After choosing 4 tens, 1 more card must be selected from the remaining 20 cards.
• Since this is the only card left to ensure a spade, we directly choose from the spades (5 spades left excluding the 10 of spades).
• Number of valid combinations: 5.

Step 4: Sum All Valid Hands

The total number of valid hands is the sum of all valid hands from each case.

$\text{Total Valid Hands} = \text{(Case 1)} + \text{(Case 2)} + \text{(Case 3)}$

$= 6 \times (\binom{20}{3} - \binom{15}{3}) + 4 \times (\binom{20}{2} - \binom{15}{2}) + 1 \times 5$

Calculation:

Let's calculate this expression:

1. Case 1 Calculation:

   • $\binom{20}{3} = 1140$
   • $\binom{15}{3} = 455$
   • $1140 - 455 = 685$
   • Case 1: $6 \times 685 = 4110$

2. Case 2 Calculation:

   • (\