in how many ways can 3 couples sit in a circular table such that no pair of husband wife are opposite to each other ?
1) 60
2) 100
3)64
4) 96
1) 60
2) 100
3)64
4) 96
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
RedeemTarget Test Prep · GMAT
Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.
TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

with Chris Peckover

with Logan Thompson
Complete access from day one. Study on your schedule.
Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.
Let the 3 couples be A and B, C and D, E and F.smanstar wrote:in how many ways can 3 couples sit in a circular table such that no pair of husband wife are opposite to each other ?
1) 60
2) 100
3)64
4) 96
Another approach:smanstar wrote:in how many ways can 3 couples sit in a circular table such that no pair of husband wife are opposite to each other ?
1) 60
2) 100
3)64
4) 96
GMATGuruNY wrote:Let the 3 couples be A and B, C and D, E and F.smanstar wrote:in how many ways can 3 couples sit in a circular table such that no pair of husband wife are opposite to each other ?
1) 60
2) 100
3)64
4) 96
With circular arrangements, where the FIRST person sits is IRRELEVANT.
All that matters is the number of ways to arrange the remaining people RELATIVE to the first person.
Case 1: C does not sit opposite A or B
Once A is seated, the number of options for B = 4. (Of the 5 remaining seats, 4 are not opposite A.)
The number of options for C = 2. (Of the 4 remaining seats, 2 are not opposite A or B.)
Number of options for D = 2. (Of the 3 remaining seats, 2 are not opposite C.)
Number of options for E = 2. (Either of the 2 remaining seats.)
Number of options for F = 1. (Only 1 seat left.)
To combine these options, we multply:
4*2*2*2*1 = 32.
Case 2: C sits opposite A or B
Once A is seated, the number of options for B = 4. (Of the 5 remaining seats, 4 are not opposite A.)
Number of options for C = 2. (Must be opposite A or B.)
Of the 3 remaining seats, one is opposite A or B.
If D occupies this seat, then C and D will be opposite A and B, forcing E and F into opposite seats.
Thus, D must occupy one of the other 2 seats.
Number of options for D = 2.
Number of options for E = 2. (Either of the 2 remaining seats.)
Number of options for F = 1. (Only 1 seat left.)
To combine these options, we multply:
4*2*2*2*1 = 32.
Total options = 32+32 = 64.
New here Create free account