In an examination, there are 8 candidates out of which 3 candidates have to appear in mathematics and the rest in different subject. In how many ways can they be seated in a row, if candidtes appearing in mathematics are not to sit together ?

Explanation

To solve the problem of seating 8 candidates in a row where 3 candidates appearing in mathematics are not to sit together, we can follow these steps:

### Step 1: Calculate the total number of candidates and their arrangement without restrictions.
We have a total of 8 candidates. The total number of ways to arrange 8 candidates in a row is given by the factorial of 8:

\[ 8! = 40320 \]

### Step 2: Calculate the arrangements where the 3 mathematics candidates sit together.
To find the arrangements where the 3 mathematics candidates sit together, we can treat these 3 candidates as a single unit or block. This means we now have 6 units to arrange: the block of 3 mathematics candidates and the 5 other candidates.

The number of ways to arrange these 6 units is:

\[ 6! = 720 \]

Within the block of mathematics candidates, the 3 candidates can be arranged among themselves in:

\[ 3! = 6 \]

Thus, the total arrangements where the mathematics candidates sit together is:

\[ 6! \times 3! = 720 \times 6 = 4320 \]

### Step 3: Calculate the arrangements where the mathematics candidates do not sit together.
To find the arrangements where the mathematics candidates do not sit together, we subtract the arrangements where they sit together from the total arrangements:

\[ \text{Arrangements where they do not sit together} = 8! - (6! \times 3!) \]

Substituting the values we calculated:

\[ = 40320 - 4320 = 36000 \]

### Final Answer
The number of ways the candidates can be seated in a row such that the candidates appearing in mathematics do not sit together is:

\[ \boxed{36000} \]