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Try this beautiful problem Based on Sitting Arrangement, useful for ISI B.Stat Entrance.

Three boys of class I, 4 boys of class II and 5 boys of class III sit

in a row. The number of ways they can sit, so that boys of the same

class sit together is

- (a) \(3!4!5!\)
- (b) \(12!/(3!4!5!)\)
- (c) \((3!)^2 4!5! \)
- (d) \(3*4!5!\)

Probability

combinatorics

Permutation

But try the problem first...

Answer: (c) \((3!)^2 4!5! \)

Source

Suggested Reading

TOMATO, Problem 120

Challenges and Thrills in Pre College Mathematics

First hint

Let us take the boys of each class as an unit.Therefore there are 3 units which can be permutated in 3! ways.

Now, class I boys can permutate among themselves in 3! ways, class II boys

in 4! And class III boys in 5! ways.

Can you now finish the problem ..........

Final Step

Therefore, total number of ways is (3!3!4!*5!) = \((3!)^24!5!\)

Therefore option (c) is the correct

- https://www.cheenta.com/octahedron-problem-amc-10a-2006-problem-24/
- https://www.youtube.com/watch?v=axSw4_SuKE8

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