Possible distributions among 3 sections (sum = 5, min 1 each):
\[(1,1,3) \quad \text{and} \quad (1,2,2)\]
Case 1 — Distribution (1, 1, 3):
\[\binom{5}{1}\binom{5}{1}\binom{5}{3} = 5 \times 5 \times 10 = 250\]
This arrangement can be done in \(\dfrac{3!}{2!} = 3\) ways, so:
\[250 \times 3 = 750\]
Case 2 — Distribution (1, 2, 2):
\[\binom{5}{1}\binom{5}{2}\binom{5}{2} = 5 \times 10 \times 10 = 500\]
This arrangement can be done in \(\dfrac{3!}{2!} = 3\) ways, so:
\[500 \times 3 = 1500\]
Total number of ways:
\[750 + 1500 = \boxed{2250}\]
✅ Answer = Option (a) 2250
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