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\frac{12!}{5! \cdot 4! \cdot 3!}

$ 12! = 479001600 $

$ 5! = 120 $, $ 4! = 24 $, $ 3! = 6 $

Denominator: $ 120 \cdot 24 \cdot 6 = 120 \cdot 144 = 17280 $

\frac{479001600}{17280} = 27720

Alternatively, step-by-step:

\frac{12!}{5!4!3!} = \binom{12}{5} \cdot \binom{7}{4} \cdot \binom{3}{3} = 792 \cdot 35 \cdot 1 = 27720

Thus, the number of distinct eating sequences is $ \boxed{27720} $.Question: Suppose $ v $ is a positive multiple of $ 5 $. If $ v^3 < 1700 $, what is the greatest possible value of $ v $?

Solution: We are given that $ v $ is a positive multiple of $ 5 $ and $ v^3 < 1700 $. We want to find the largest such $ v $.

Start by testing multiples of $ 5 $:

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