The number of ways that 5 student can be made to sit in a row so that the tallest and shortest may not come together is

Options

A. 48
B. 24
C. 72
D. 120

Answer

To solve the problem of how many ways 5 students can be arranged in a row such that the tallest and shortest students do not sit together, we can follow these steps:

### Step 1: Calculate the total arrangements without restrictions
First, we calculate the total number of arrangements of 5 students without any restrictions. The formula for the number of permutations of n distinct objects is given by \( n! \) (n factorial).

\[ \text{Total arrangements} = 5! = 120 \]

### Step 2: Calculate the arrangements where the tallest and shortest sit together
Next, we treat the tallest and shortest students as a single unit or block. This means we will have 4 units to arrange: the block of the tallest and shortest, and the other 3 students.

The number of ways to arrange these 4 units is:

\[ \text{Arrangements of units} = 4! = 24 \]

Since the tallest and shortest can be arranged within their block in 2 ways (Tallest first or Shortest first), we multiply by 2:

\[ \text{Arrangements with tallest and shortest together} = 4! \times 2 = 24 \times 2 = 48 \]

### Step 3: Calculate the arrangements where the tallest and shortest do not sit together
To find the number of arrangements where the tallest and shortest do not sit together, we subtract the arrangements where they sit together from the total arrangements:

\[ \text{Arrangements where tallest and shortest do not sit together} = \text{Total arrangements} - \text{Arrangements with tallest and shortest together} \]

\[ = 120 - 48 = 72 \]

### Final Answer
The number of ways that 5 students can be arranged in a row such that the tallest and shortest do not sit together is **72**.

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