In how many ways 4 persons can occupy 10 chairs in a row, if no two sit on adjacent chairs?

Answer

To solve the problem of how many ways 4 persons can occupy 10 chairs in a row such that no two persons sit on adjacent chairs, we can follow these steps:

### Step 1: Identify the total number of chairs and persons
We have a total of 10 chairs and we need to seat 4 persons.

### Step 2: Determine the empty chairs
Since no two persons can sit next to each other, we need to ensure that there is at least one empty chair between any two persons.

### Step 3: Place the empty chairs
If we place 4 persons, they will occupy 4 chairs. To ensure that no two persons are adjacent, we will need to place at least 3 empty chairs between them. This means that we will have:

- 4 chairs occupied by persons
- 3 chairs as mandatory empty chairs between them

This accounts for 7 chairs (4 occupied + 3 mandatory empty).

### Step 4: Calculate remaining empty chairs
Since we have 10 chairs in total, after placing 7 chairs (4 occupied + 3 mandatory empty), we have:

10 - 7 = 3 empty chairs remaining.

### Step 5: Determine the arrangement of chairs
Now we need to arrange these 3 remaining empty chairs. We can visualize the arrangement as follows:

- We have 4 persons (P) and 3 mandatory empty chairs (E) already placed, which can be represented as:

E P E P E P E P E

Now, we have 3 additional empty chairs that can be placed in the gaps created by the arrangement of persons and mandatory empty chairs.

### Step 6: Calculate the total gaps
The arrangement creates gaps where the additional empty chairs can be placed. The gaps are:

- Before the first person
- Between the first and second person
- Between the second and third person
- Between the third and fourth person
- After the fourth person

This gives us a total of 5 gaps.

### Step 7: Distribute the remaining empty chairs
We need to distribute the 3 remaining empty chairs into these 5 gaps. This can be done using the "stars and bars" theorem, which states that the number of ways to distribute \(n\) indistinguishable objects (empty chairs) into \(k\) distinguishable boxes (gaps) is given by:

\[ \text{Number of ways} = \binom{n + k - 1}{k - 1} \]

In our case, \(n = 3\) (remaining empty chairs) and \(k = 5\) (gaps). Thus, we have:

\[ \text{Number of ways} = \binom{3 + 5 - 1}{5 - 1} = \binom{7}{4} \]

### Step 8: Calculate the arrangements of persons
Now, we also need to arrange the 4 persons in the selected chairs. The number of ways to arrange 4 persons is given by \(4!\):

\[ 4! = 24 \]

### Step 9: Combine the results
The total number of arrangements is the product of the ways to choose the gaps and the arrangements of the persons:

\[ \text{Total arrangements} = \binom{7}{4} \times 4! = 35 \times 24 = 840 \]

### Final Answer
Thus, the total number of ways 4 persons can occupy 10 chairs in a row, ensuring that no two sit on adjacent chairs, is **840**.

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