Question on permutation?

Question

there are 8 students appearing in examination,of which 3 have to appear in maths paper and remaining 5 in different subjects,in how many ways can made to sit in a row if the candidates of maths cant sit next to each other.

Answer 1

The number of combinations is the product of (1) the number of ways to seat the maths candidates, (2) the number of ways to seat the other candidates, and (3) the number of ways to place the maths candidates among the other candidates so that they do not sit next to each other.

The candidates of maths can sit in 3! = 6 different ways, and the other candidates can sit in in 5! = 120 different ways.

We ensure that the candidates of maths do not sit next to each other in the following way.  Imagine that you have seated the 5 other candidates.  We can seat any of the maths candidates at either end or between any two of the other candidates.  This gives us 6 seating options to choose from (left end, between 1&2, between 2&3, between 3&4, between 4&5, right end), and as long as we don't give two different students the same seating option, they won't sit next to each other.

There are (6 choose 3) = (6!)/(3!3!) = 20 different ways to choose seating for three students among the 5 other candidates so that the three students do not sit next to each other.

Now all that is left is the calculation:

6 X 120 X 20 = 14400

There are 14,400 ways to arrange the students so that the candidates of maths do not sit next to each other.