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
Probability
combinatorics
Permutation
Answer: (c) \((3!)^2 4!5! \)
TOMATO, Problem 120
Challenges and Thrills in Pre College Mathematics
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 ..........
Therefore, total number of ways is (3!3!4!*5!) = \((3!)^24!5!\)
Therefore option (c) is the correct
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
