In an examination, there are 8 candidates out of which 3 candidates have to
Question:

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?

 

Solution:

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 $40320-4320=36000$.

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.