In how many ways 10 passengers can board three buses if no bus remains empty?

Answer

To solve the problem of how many ways 10 passengers can board three buses such that no bus remains empty, we can use the principle of inclusion-exclusion. Here’s the step-by-step solution:

### Step 1: Calculate Total Arrangements Without Restrictions
First, we calculate the total number of ways to assign 10 passengers to 3 buses without any restrictions. Each passenger has 3 choices (Bus 1, Bus 2, or Bus 3).

\[ \text{Total arrangements} = 3^{10} \]

### Step 2: Subtract Arrangements with At Least One Empty Bus
Next, we need to subtract the cases where at least one bus is empty. We can do this using the principle of inclusion-exclusion.

#### Step 2.1: Choose One Bus to be Empty
If we choose one bus to be empty, we can select which bus to empty in \( \binom{3}{1} \) ways (3 choose 1). The remaining 10 passengers can then be assigned to the 2 remaining buses.

\[ \text{Ways with one bus empty} = \binom{3}{1} \cdot 2^{10} \]

#### Step 2.2: Add Back Arrangements with Two Buses Empty
However, when we subtract the cases with one bus empty, we have also subtracted the cases where two buses are empty twice. We need to add these cases back.

If two buses are empty, there is only one bus left for all 10 passengers. The number of ways to choose which two buses are empty is \( \binom{3}{2} \) and all passengers must go to the remaining bus.

\[ \text{Ways with two buses empty} = \binom{3}{2} \cdot 1^{10} \]

### Step 3: Combine the Results
Now we can combine these results using the inclusion-exclusion principle:

\[ \text{Valid arrangements} = 3^{10} - \binom{3}{1} \cdot 2^{10} + \binom{3}{2} \cdot 1^{10} \]

### Step 4: Calculate the Values
Now we substitute the values:

- \( 3^{10} = 59049 \)
- \( \binom{3}{1} = 3 \)
- \( 2^{10} = 1024 \)
- \( \binom{3}{2} = 3 \)
- \( 1^{10} = 1 \)

Now substituting these values into the equation:

\[ \text{Valid arrangements} = 59049 - 3 \cdot 1024 + 3 \cdot 1 \]

Calculating this gives:

\[ \text{Valid arrangements} = 59049 - 3072 + 3 = 55980 \]

### Final Answer
Thus, the total number of ways 10 passengers can board 3 buses such that no bus remains empty is:

\[ \boxed{55980} \]