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Try this Merry-go-round Problem based on the combinatorics from TOMATO useful for ISI B.Stat Entrance.

Four married couples are to be seated in a merry-go-round with 8 identical seats. In how many ways can they be seated so that:

i) males and females seat alternately, and

ii) no husband seats adjacent to his wife?

- 8
- 12
- 16
- 20

combinatorics

probability

Number theory

But try the problem first...

Answer: \(12\)

Source

Suggested Reading

TOMATO, Problem 104

Challenges and Thrills in Pre College Mathematics

First hint

There are \(8\) persons......\(W_1,W_2,W_3,W_4,M_1,M_2,M_3,M_4\).Given that males and females seat alternately & no husband seats adjacent to his wife.Let us assume that .\(W_1\) is the wife of \(M_1\),\(W_2\) is the wife of \(M_3\) and the similar for others.....

Therefore \(M_1\) can not be seat beside or after \(W_1\).similar for others.can you draw a circular form ...?

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

Second Hint

\(W_1\) can sit in two seats either in the seat in left side figure or in the seat in

right side figure. In left side figure when \(W_1\) is given seat then \(W_4\) can sit in

one seat only as shown and accordingly \(W_2\) and \(W_3\) can also take only one

seat. Similarly, right side figure also reveals one possible way to seat. So

there are two ways to seat for every combination of Men

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

Final Step

Now, Men can arrange themselves in (4 – 1)! = 6 ways. So number of ways = \(2 \times 6 \)= \(12\).

- https://www.cheenta.com/problem-based-on-lcm-amc-8-2016-problem-20/
- https://www.youtube.com/watch?v=V01neV8qmh4

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