How many ways are there to seat 3 couples at a round table such that each person does not sit opposite to his/her spouse?
^3 =2*2*2=8. Hence the total number of arrangements are : 6*8=48. Hence, the answer is 48.
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 four married couples sit on a park bench if?
384 ways
Therefore, if there are 4 married couples sitting together, there are 384 ways of seating arrangement.
How many arrangements are there where the couples are seated 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.
How many seats can a wife sit next to her husband?
That leaves 2/5 seats where her husband can sit next to her. Second wife, say, can take any of the four remaining seats. There is then only 1 seat out of the remaining 3 where her husband can sit next to her AND leave two empty adjacent seats for the last couple.
How many couples can be seated in $2^3$ ways?
There are three couples, so each one of them can be seated in $2^3$ ways as couple. Now the three couples are three people and they can be placed in around the circular table in $2!$ ways. Now the required probability $= \\frac{2^3.2}{5!} = \\frac{2}{15}$
How many couples are sitting randomly round a 6-seater table?
There are 3 couples sitting randomly round a 6-seater circular table. What is the probability that all the husbands and wives sit next to each other? My attempt: First wife, say, takes any of the…
How do you permute a table with 3 couples?
Fix each couple as one. They can be permuted in 2 ways. There are three couples, so each one of them can be seated in $2^3$ ways as couple. Now the three couples are three people and they can be placed in around the circular table in $2!$ ways.