In an examination, there are 8 candidates out of which 3 candidates ha

1.	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?
Answer» 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 arrangement are P(6,6) × 3! = \(\frac{6!}{(6-6)!}\times3!\) = \(\frac{6!}{0!}\times3!\) = \(\frac{720}{1}\times6\) = 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.

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?

Answer»

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 arrangement are

P(6,6) × 3! = \(\frac{6!}{(6-6)!}\times3!\)

= \(\frac{6!}{0!}\times3!\) = \(\frac{720}{1}\times6\) = 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.