# If three couples are seated around a circular table...?

If three couples are seated around a circular table, what is the probability that no wife and husband are beside one another?

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let's do it for FOUR couples using inclusion-exclusion for a *numbered* circle (it won't matter, as probability has been asked for)

the terms in sequence will represent

1. choose couples: 4c1,4c2,4c3,4c4

2. place them: (8/1)*6c0=8, (8/2)*5c1=20, (8/3)*4c2=16, (8/4)*3c3=2

[ (a couple "train" starting from any of 8 points) ÷ ( # of couples to account for rotational invariance), and "singles" placed between trains using "stars & bars" ]

3. arrange them: 1!,2!,3!,4!

4. flip spouses: 2¹,2²,2³,2⁴

5. permute rest: 6!,4!,2!,0!

applying inclusion-exclusion,

# of ways = 8! - 4*8*1*2*6! + 6*20*2*4*4! -4*16*3!*8*2! +1*2*4!*16*1 = 11,904

Pr = 11,904/8! = 31/105, ≈ 0.295238 <-------

4 couples were used to show the generality of this method

for 3 couples, you can do a similar calculation to get 192 ways,

Pr = 192/6! = 4/15 <------

the direct formula that emerges is

Pr = 1 + (1/(2n)!)* sum (-1)^k *C(n,k) *(2n/k)*C(2n-k-1,k-1) *k! * 2^k*(2n-2k)!, k = 1 to n

the computation of Pr = 4/15 for n = 3 can be seen at

http://www.wolframalpha.com/input/?i=1+%2B+%281%2F...

• Put any of the wives in a seat, any seat. Now her husband can be placed either

two seats clockwise or counterclockwise, or directly opposite, i.e., 3 chairs around.

Consider these three cases separately:

1.) two seats clockwise: now place one of the four remaining people in between

the already seated couple. Then their spouse will have to go directly opposite

so as to keep the final couple apart. This final couple can be arranged in two ways,

so there are 4*2 = 8 arrangements in this case.

2.) two seats counterclockwise: same process as in 1.), giving 8 more arrangements.

3.) directly opposite: now place one of the four remaining people one seat clockwise

from the first wife seated. Then one spouse from the remaining couple can be

placed one seat clockwise. In the remaining two seats, the last two people are not

spouses so can be arranged in 2 ways, giving us 4*2*2 = 16 more arrangements.

This gives us a total of 8 + 8 + 16 = 32 arrangements where no wife and husband

sit next to each other. Since there are 5! = 120 ways of arranging 6 people around

a circular table without restrictions, this gives us a probability of 32/120 = 4/15,

or 26.667%, that no wife and husband sit next to each other.

• First, imagine that the each and every human being is being seated in a line, no longer at a round table. 3! = 6 techniques to reserve the three couples that should be saved at the same time, e.g. ABC, ACB, etc. interior each and every couple, there are 2 techniques to reserve the individuals: husband/spouse, or spouse/husband. for 3 couples, that is 2^3 = 8 opportunities. There are 2 human beings left to position. 4 places to position the first one: - To the left of the first couple, e.g. xABC - between the first and 2d couples, e.g. AxBC - between the 2d and third couples, e.g. ABxC - To the right of the third couple, e.g. ABCx Now there are 5 techniques to position the in basic terms right human being, e.g. yxABC, xyABC, xAyBC, xAByC, xABCy finished opportunities: 6 * 8 * 4 * 5 = 960 notwithstanding, because that they're at a round table, you may want to assert that rotating the table makes no distinction, i.e. a format of 12345678 is an similar as 23456781 and 34567812, etc. hence, divide the above huge type by technique of 8 to get one hundred twenty.

• N=total arrangements= 6! (6 people arranged in order)

A_1 (A subscript 1) = couple one together = 5! (4 people and one couple arranged)

A_2 = couple two together = 5!

A_3 = couple three together = 5!

A_4 = couple one together AND couple two together = 4! (2 people and two couples arranged)

A_5 = couple one together and couple three together = 4!

A_6 = couple two together and couple three together = 4!

A_7 = couple one together and couple two together and couple three together = 3! (3 couples arranged)

# arrangements that meet requirements are: N -(A_1 + A_2 + A_3) +(A_4 + A_5 + A_6) -A_7

= 6! -3*5! +3*4! -3! = 426.

Probability: 426/N= 426/6! = .59167 or 59%

Source(s): I worked this out based on "Applied Combinatorics" 5th Edition by Alan Tucker.