<p>A train timetable must be compiled for various days of the week so that two trains twice a day depart for three days, one train daily for two days, and three trains once a day for two days. How many different timetables can be compiled?</p>
Step-by-Step Solution
Key Concept: This is a multinomial arrangement problem where we must partition 7 days into three distinct groups (2 days with 2 trains, 2 days with 1 train, 2 days with 3 trains) and then arrange the specific train counts within each partition.
<p><strong>Step 1:</strong> Identify the partition structure. We have 7 days total: 2 days with 2 trains each, 2 days with 1 train each, and 2 days with 3 trains each, with 1 day remaining (or reinterpret as needing to assign 2+2+1+1+3+3 = 12 trains across 7 days - actually, recount: 2 days × 2 trains + 2 days × 1 train + 2 days × 3 trains = 4+2+6=12 train slots across 6 days).</p><p><strong>Step 2:</strong> Select which days receive which frequency. The number of ways to partition 7 days into groups of size 2, 2, 2, 1 (two groups of 2 days with 2 trains, two groups of 2 days with different train counts, and 1 remaining day) is the multinomial coefficient: $\frac{7!}{2! \cdot 2! \cdot 2! \cdot 1!}$</p><p><strong>Step 3:</strong> Reconsider: Select 2 days for twice-daily trains: $\binom{7}{2}$. From remaining 5, select 2 days for once-daily: $\binom{5}{2}$. From remaining 3, select 2 days for thrice-daily: $\binom{3}{2}$. This gives: $\binom{7}{2} \cdot \binom{5}{2} \cdot \binom{3}{2} = 21 \cdot 10 \cdot 3 = 630$</p><p><strong>Step 4:</strong> If trains are distinguishable, multiply by the number of ways to arrange them: $2! \cdot 1! \cdot 3!$ for each train set, but the problem structure suggests the answer is the selection itself.</p><p>∴ Answer: A (630 or 7!/(2!·2!·2!·1!) = 1260, depending on distinguishability assumption)</p>
Correct Answer: A