Six text books numbered 1, 2, 3, 4, 5 and 6 are arranged at random. What is the probability that the text books 2 and 3 will occupy consecutive places?

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. `(1)/(6)`

Answer

To find the probability that textbooks 2 and 3 occupy consecutive places when six textbooks numbered 1, 2, 3, 4, 5, and 6 are arranged at random, we can follow these steps:

### Step 1: Calculate the Total Arrangements of the Textbooks
The total number of ways to arrange 6 textbooks is given by the factorial of the number of textbooks.

\[ \text{Total arrangements} = 6! = 720 \]

### Step 2: Treat Textbooks 2 and 3 as a Single Unit
Since we want textbooks 2 and 3 to be together, we can treat them as a single unit or block. This means we now have 5 units to arrange: the block (2 and 3), and the other textbooks 1, 4, 5, and 6.

\[ \text{Units to arrange} = \{(2,3), 1, 4, 5, 6\} \]

### Step 3: Calculate the Arrangements of the Units
The number of ways to arrange these 5 units is given by the factorial of the number of units.

\[ \text{Arrangements of units} = 5! = 120 \]

### Step 4: Arrange Textbooks 2 and 3 Within Their Block
Within the block (2 and 3), the two textbooks can be arranged in 2 different ways (2 can be first or 3 can be first).

\[ \text{Arrangements of (2, 3)} = 2! = 2 \]

### Step 5: Calculate the Total Favorable Arrangements
Now, we can find the total number of arrangements where textbooks 2 and 3 are together by multiplying the arrangements of the units by the arrangements within the block.

\[ \text{Total favorable arrangements} = 5! \times 2! = 120 \times 2 = 240 \]

### Step 6: Calculate the Probability
Finally, we can calculate the probability that textbooks 2 and 3 occupy consecutive places by dividing the total favorable arrangements by the total arrangements.

\[ \text{Probability} = \frac{\text{Total favorable arrangements}}{\text{Total arrangements}} = \frac{240}{720} = \frac{1}{3} \]

Thus, the probability that textbooks 2 and 3 will occupy consecutive places is:

\[ \boxed{\frac{1}{3}} \]