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
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circular arrangement
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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.
Last edited by GMATGuruNY on Mon Sep 10, 2012 6:18 pm, edited 1 time in total.
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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
Good arrangements = total possible arrangements - bad arrangements.
In a bad arrangement, at least 1 couple is in opposite seats.
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.
Let the 3 couples be A and B, C and D, E and F.
Total possible arrangements:
Once A is seated, the number of ways to arrange the 5 remaining people RELATIVE to A = 5! = 120.
Bad arrangement 1: Every couple in opposite seats
Once A is seated, the number of options B = 1. (Must be opposite B.)
Number of options for C = 4. (Any of the 4 remaining seats.)
Number of options for D = 1. (Must be 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 multiply:
1*4*1*2*1 = 8.
Bad arrangement 2: Exactly two couples in opposite seats
Not possible: If two couples are in opposite seats, then the third couple must also be in opposite seats.
Bad arrangement 3: Exactly one couple in opposite seats
Let A and B be the one couple in opposite seats.
Once A and B are seated, the number of options for C = 4. (Any of the 4 remaining seats.)
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 two remaining seats.)
Number of options for F = 1. (Only 1 seat left.)
To combine these options, we multiply:
4*2*2*1 = 16.
Since the one couple sitting in opposite seats could be AB, CD, or EF, we multiply by 3:
3*16 = 48.
Good arrangements = 120-8-48 = 64.
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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.
Hi Mitch
Can you please explain , why we considered only the cases for C sitting opposite to A or B ??
Should we not find it for all the people in D, E, F and then add them ??
