A ballroom contains four married couples (four men and their four wives). The ge

Question

A ballroom contains four married couples (four men and their four wives). The gentlemen are to be paired up randomly with the ladies, one man to each woman. After one dance, the eight people are to be separated, except for the married couples that happen to be together by chance. The separated people will then be paired up again randomly for the second dance, again one man to each woman.

(a) Find the probability that exactly two married couples will be dancing together on the second dance.

(b) Suppose that exactly one married couple are dancing together on the second dance. What is the probability that no man was dancing with his wife on the first dance?

Explanation / Answer

this is a problem of derangements & partial derangements
a derangement of n objects Dn is one where none of them are in their proper place.
in a partial derangement, only some of them aren't in their proper place

in this q, let us consider one sex as fixed, & one as floating
since n is small, we shall simply enumerate w/o going into formulas
if you want formulas, see sources cited !

D4 = 9: BADC BCDA BDAC CADB CDAB CDBA DABC DCAB DCBA
D3 = 2: BCA CAB
D2 = 1: BA

what about partial derangements ?
suppose 2 are deranged out of 4
D2 = 1, but ANY 2 could be deranged,
so ways would be 1*4C2 = 6

let us make a matrix of possibilities to ease calculations
e,g when n = 4, full derangement is 9, 2 deranged is 6
note that you can never have just 1 deranged !

n | # deranged 4 .. 3 .. 2
---|---------- ------------ -------
4 |. . . . . . . . . 9 .. 8 .. 6

3 |. . . . . . . . . . . . 2 .. 3

2 |. . . . . . . . . . . . . . 1

we are now in a good position to attack the problem !

(a)
2 out of 4 deranged in round 2 can happen in many ways

4 deranged in round 1, 2 of 4 in round 2
----------- --------- ------- ----------
probability = [9/4!]*[6/4!]= 54/576

3 deranged in round 1, 2 of 3 in round 2
------------ ---------- ----------------
probability = [8/4!]*[3/3!] = 96/576

2 deranged in round 1, 2 of 2 in round 2
--------- ----------- ------------------
probability = [6/4!]*[1/2!] = 72/576

so reqd. probability = 222/576 = 111/288 = 30.3854

(b)
we want P[4 deranged in round 1 | 3 deranged in round 2]

P[4 in rd 1 & 3 of 4 in rd 2] = [9/4!]*[8/4!] = 72/576 = A

P[3 in rd 1 & 3 of 3 in rd 2] = [8/4!]*[2/3!] = 64/576

P[3 deranged in rd 2] = sum of the above = 136/576 = B

reqd.probability = A/B = 72/136 = 9/17 = 0.5294

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