Questions

How many ways can three couples be arranged in a line such that each husband is ahead of his wife?

How many ways can three couples be arranged in a line such that each husband is ahead of his wife?

I tried to solve this, as 3 couples are there and they should be together, number of permutations are 3! and as it is fixed that each husband will be ahead of his wife, no further permutation is required for husband and wife. Thus, my answer is 3! =6, but it is nowhere close to any of the option.

How many ways can they be seated if each couple is to sit together?

Therefore, there are 40320 ways that the people can be seated when there is no restriction on the seating arrangement. The possible ways = 7! × 2! Therefore, if persons A and B must sit next to each other there are 10080 ways of seating arrangement.

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How many ways 5 couples can sit in a linear row such that husband and wife are not sitting adjacent to each other #pnc?

In how many ways can five couples be seated in a row if each couple must be seated together? – Quora. Treating each couple as 1 unit, then the 5 couples can be seated 5! = 120 ways. Since spouses can switch seats but still sit together, each couple be seated in 2!

What are the number of ways of making 5 couples stand in a line?

Five people can line up in 5! = 120 ways.

How many ways can 3 people sit?

So there are two answers: There are 3! = 6 different ways of placing these three people in three distinct chairs. However, it we decide to consider rotated arrangements as basically the same, then there are only 2 ways.

How many ways can 4 married couples seat themselves around a circular table if spouses sit opposite each other?

Below are two specific examples. Ex2. Mom and dad and their 6 children (3 boys and 3 girls) are to be seated at a table. How many ways can this be done if mom and dad sit together and the males and females alternate?

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How many ways are there to seat n married couples?

There are 720 different permutations of couples.

How many ways are there to select a subcommittee of 3 members from among a committee of 10?

(10−3)! = 120.