Q. 253.8( 4 Votes )

In an examination

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.


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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

Four letters E, KRD Sharma - Mathematics

. There are 6 iteRD Sharma - Mathematics

How many permutatRD Sharma - Mathematics

Find the total nuRD Sharma - Mathematics

In how many ways RD Sharma - Mathematics

From among the 36RD Sharma - Mathematics

How many differenRD Sharma - Mathematics

How many three-diRD Sharma - Mathematics

In how many ways RD Sharma - Mathematics

How many three-diRD Sharma - Mathematics