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04 Aug 2025, 23:31
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Seven men and five women have to sit around a circular table so that no 2 women are together.
In how many different ways can this be done?
There are (7-1)! = 6!=720 to get seven men seated around a circular table
Since there are no 2 women sitting together, there is 1 or more men sitting between 2 women. There are 7 vacancies in between 7 men to be filled by 5 women. Number of ways to do that = 7C5 *5! = 21*120
Total number of ways in which seven men and five women have to sit around a circular table so that no 2 women are together = 6!*7C5*5! = 1,814,400
IMO D
In how many different ways can this be done?
There are (7-1)! = 6!=720 to get seven men seated around a circular table
Since there are no 2 women sitting together, there is 1 or more men sitting between 2 women. There are 7 vacancies in between 7 men to be filled by 5 women. Number of ways to do that = 7C5 *5! = 21*120
Total number of ways in which seven men and five women have to sit around a circular table so that no 2 women are together = 6!*7C5*5! = 1,814,400
IMO D
06 Aug 2025, 16:53
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I solved by making total no.of possible ways -(no. of ways not required)
For this no. of ways not required would be women always seated together -> If I consider them as one single unit, and then arrange them with the total men would this not be (7+1-1)! 5! = 7!5!
Answer =11! - (7!5!)
Can someone please let me know why this would be wrong?
For this no. of ways not required would be women always seated together -> If I consider them as one single unit, and then arrange them with the total men would this not be (7+1-1)! 5! = 7!5!
Answer =11! - (7!5!)
Can someone please let me know why this would be wrong?
07 Aug 2025, 00:06
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BDNN
Your approach subtracts only the case where all 5 women sit together, but the restriction is stricter, no two women can sit together. There are many invalid cases where women sit in smaller groups (like 2 or 3 together), which your subtraction doesn’t cover. So 11! - 7!5! doesn’t remove all the unwanted arrangements.
