((5χ/3)! * (χ/3)!)/((4χ/3) * (2χ/3)!)
asked by guest on Jan 29, 2025 at 12:09 pm
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MathBot Answer:
(5χ3)!⋅(χ3)!4χ3⋅(2χ3)!=3(χ3)!(5χ3)!4χ(2χ3)!\frac{\left( \frac{5 χ}{3} \right)! \cdot \left( \frac{χ}{3} \right)!}{\frac{4 χ}{3} \cdot \left( \frac{2 χ}{3} \right)!} = \frac{3 \left(\frac{χ}{3}\right)! \left(\frac{5 χ}{3}\right)!}{4 χ \left(\frac{2 χ}{3}\right)!}34χ⋅(32χ)!(35χ)!⋅(3χ)!=4χ(32χ)!3(3χ)!(35χ)!