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To solve this, you have to break it down into the same "Select then Arrange" flow:
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Selection ($C$): * We need 3 Physics books. Since P1 is mandatory, we only need to choose 2 more from the remaining 4. Calculation: $\binom{4}{2} = 6$.
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We need 2 Math books from 4. Calculation: $\binom{4}{2} = 6$.
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Total selection combinations: $6 \times 6 = 36$.
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Arrangement ($P$): * We have 5 books (3 Physics, 2 Math).
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The "Gap Method" is best for the constraint "Math books not together."
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First, arrange the 3 Physics books: $3! = 6$ ways.
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This creates 4 possible gaps ( _ P _ P _ P _ ).
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We must place the 2 Math books into these 4 gaps: $^4P_2 = 12$ ways.
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Total arrangements per selection: $6 \times 12 = 72$.
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Final Calculation:
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$36 \text{ (selections)} \times 72 \text{ (arrangements)} = 2592$.
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Correct Answer: D (1200 or more)
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