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?
combinatorics
probability
Number theory
Answer: \(12\)
TOMATO, Problem 104
Challenges and Thrills in Pre College Mathematics
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 ..........
\(W_1\) 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........
Now, Men can arrange themselves in (4 – 1)! = 6 ways. So number of ways = \(2 \times 6 \)= \(12\).
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