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 $\mathrm{P}(6,6)$ 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 arrangement are
$P(6,6) \times 3 !=\frac{6 !}{(6-6) !} \times 3 !$
$=\frac{6 !}{0 !} \times 3 !=\frac{720}{1} \times 6=4320$
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 $40320-4320=36000$.