In an examination, there are 8 candidates out of which 3 candidates have to appear in mathematics and the rest in different subjects. In how many ways can they are seated in a row if candidates appearing in mathematics are not to sit together?
Candidates in mathematics are not sitting together = total ways – the
Students are appearing for mathematic sit together.
The total number of arrangements of 8 students is 8! = 40320
When students giving mathematics exam sit together, then consider
them as a group.
Therefore, 6 groups can be arranged in P(6,6) ways.
The group of 3 can also be arranged in 3! Ways.
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n,r) = n!/(n-r)!
Therefore, total arrangments are
P(6,6) × 3! = 
×3!
= 
×3! = 
×6 = 4320.
The total number of possibilities when all the students giving
mathematics exam sits together is 4320 ways.
Therefore, number of ways in which candidates appearing
mathematics exam is 40320 – 4320 = 36000.