Answer: 604806048060480 ways.

Understanding the problem

You have:

  • 7 periods in a day.
  • 6 different subjects.
  • Each subject must get at least one period.
  • One of the subjects will therefore get 2 periods , and the remaining five get 1 period each.

We are only arranging periods within a single day , not across multiple days.

Step-by-step solution

1. Choose which subject gets 2 periods

Out of the 6 subjects, choose 1 subject to take 2 periods.

Ways=(61)=6\text{Ways}=\binom{6}{1}=6Ways=(16​)=6

This is because any one of the 6 subjects could be the one that appears twice.

2. Arrange the 7 periods

Now we have a multiset of “period-entries” for the day:

  • The chosen subject appears 2 times.
  • Each of the remaining 5 subjects appears 1 time.

So, in total, we are arranging 7 “slots”: 2 identical copies of one subject, and 5 distinct others. Number of distinct arrangements of these 7 positions is:

7!2!\frac{7!}{2!}2!7!​

Because:

  • 7!7!7! counts all permutations if all 7 were distinct.
  • Divide by 2!2!2! to correct for overcounting the two identical periods of the repeated subject.

Compute:

7!=5040,2!=2,7!2!=50402=25207!=5040,\quad 2!=2,\quad \frac{7!}{2!}=\frac{5040}{2}=25207!=5040,2!=2,2!7!​=25040​=2520

3. Multiply both choices

Total number of ways:

Total ways=(61)×7!2!=6×2520=15120\text{Total ways}=\binom{6}{1}\times \frac{7!}{2!} =6\times 2520 =15120Total ways=(16​)×2!7!​=6×2520=15120

But note: this counts each arrangement once per choice of the repeated subject. However, in this formulation, we already baked in the choice of which subject is repeated; we do not need any further division or multiplication. Let’s check logic with a small analogy to avoid confusion:

  • If there were 3 periods and 2 subjects A, B, each at least once:
    • One subject is repeated: either A or B → 2 choices.
    • If A is repeated: A, A, B arranged in 3!/2!=33!/2!=33!/2!=3 ways.
    • If B is repeated: B, B, A arranged in 3!/2!=33!/2!=33!/2!=3 ways.
    • Total = 2×3=62\times 3=62×3=6 arrangements.
    • You can list them to confirm: AAB, ABA, BAA, BBA, BAB, ABB → 6.

So the same pattern holds here. Thus the correct total:

15120\boxed{15120}15120​

Final checked answer

There are 151201512015120 distinct ways to organize 6 subjects in 7 periods in a day so that every subject gets at least one period.