On five chairs arranged in a row, five persons A, B, C, D and E are to be seats in such a way that B and D always sit together (side by side). In how many different ways can it be done?

Options

A. 120
B. 48
C. 60
D. 24

Answer

To solve the problem of seating five persons A, B, C, D, and E on five chairs arranged in a row, with the condition that B and D must always sit together, we can follow these steps:

### Step 1: Treat B and D as a single unit
Since B and D must sit together, we can consider them as one single unit or block. We can denote this block as (BD) or (DB) depending on their arrangement.

### Step 2: Count the units
Now, instead of five individuals (A, B, C, D, E), we have four units to arrange:
1. The block (BD) or (DB)
2. Person A
3. Person C
4. Person E

So, we have four units: (BD), A, C, and E.

### Step 3: Arrange the units
The number of ways to arrange these four units is given by the factorial of the number of units. Therefore, the number of arrangements is:
\[ 4! = 24 \]

### Step 4: Arrange B and D within their block
Within the block (BD), B and D can be arranged in two ways: either B sits first and D second (BD) or D sits first and B second (DB).

### Step 5: Calculate total arrangements
Since there are 2 ways to arrange B and D within their block, we multiply the arrangements of the units by the arrangements of B and D:
\[ \text{Total arrangements} = 4! \times 2 = 24 \times 2 = 48 \]

Thus, the total number of different ways to seat A, B, C, D, and E such that B and D are always sitting together is **48**.

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