Riffing off prony's answer, which I think is a little confusing. Here are the possibilities for each card:

• Card 1: 52 cards
• Card 2: 3, since it must match the suite of Card 1
• Card 3: 48, since we can't match the suite of Card 1
• Card 4: 3, since we can't match the suite of Card 3
• Card 5: 44, since we can't match the suite of the other cards

This will give us all orderings of the form XXYYZ. We then notice two issues that we are double counting:

1. Cards 1 and 2 can be interchanged (XX). (2!)
2. Cards 3 and 4 can be interchanged (YY). (2!)
3. Cards 1 and 2 can collectively be interchanged with Cards 3 and 4 (XX with YY). (2!)

So we have distinct, unordered $(52 \times 3 \times 48 \times 3 \times 44)/(2! 2! 2!)$ ways. Dividing this by the number of combinations ${52 \choose 5}$ yields our answer.

Riffing off prony's answer, which I think is a little confusing. Here are the possibilities for each card:

• Card 1: 52 cards
• Card 2: 3, since it must match the value of Card 1
• Card 3: 48, since we can't match the value of Card 1
• Card 4: 3, since we can't match the value of Card 3
• Card 5: 44, since we can't match the value of the other cards

This will give us all orderings of the form XXYYZ. We then notice two issues that we are double counting:

1. Cards 1 and 2 can be interchanged (XX). (2!)
2. Cards 3 and 4 can be interchanged (YY). (2!)
3. Cards 1 and 2 can collectively be interchanged with Cards 3 and 4 (XX with YY). (2!)

So we have distinct, unordered $(52 \times 3 \times 48 \times 3 \times 44)/(2! 2! 2!)$ ways. Dividing this by the number of combinations ${52 \choose 5}$ yields our answer.