Now, \( a + b = 15 \):
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- So \( a + b = 6 \) or \( a + b = 15 \) (since \( a \geq 1 \), \( b \leq 9 \), maximum \( a + b = 9 + 9 = 18 \), but we'll check both).
- First, count solutions to \( a + b = 6 \), \( a \geq 1 \), \( b \geq 0 \):
- \( a \) ranges from 1 to 6 → \( b = 6 - a \geq 0 \), so 6 solutions.
- \( a \geq 6 \) (since \( b \leq 9 \)), and \( a \leq 9 \). So \( a = 6,7,8,9 \):
- \( a = 6 \), \( b = 9 \)
- \( a = 7 \), \( b = 8 \)
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