Probability of Married Couple Having Nonadjacent Desks

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SUMMARY

The probability that a married couple will have nonadjacent desks when randomly assigned among six desks is definitively 2/3. This conclusion is reached by first calculating the probability that the couple sits adjacent to each other, which is found to be 1/3. The calculation involves treating the married couple as a single unit, resulting in 2(5!) arrangements for adjacent seating out of a total of 6! arrangements. Thus, the final probability for nonadjacent seating is 1 - 1/3 = 2/3.

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mr_coffee
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Hello everyone...im feeling lost on this question. The answer is 2/3. But i don't know how they are getting it.

Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks? (Hint: First find the probability that the coup will have adjacent desks, and then subtract this number from 1.)

Well I think i figured out how many ways the 2 married couple will end up in adjacent desks by the following:

I let A, B, C, D, E, F stand for the 6 employee's, I'm going to assume A and B are the married couple.
So...
[number of ways to place 6 together keeping A and B side by side] = [Number of ways to arrange [A B] C D E F ] + [ number of ways to arrange [B A] C D E F] = 5! + 5! = 120 + 120 = 240 ways to arrange the 6 employee's keeping A and B together (the married couple).

Now i know the total number of ways to place the 6 people together wolud be 6! = 720. So the probability that the couple will have adjacent desks is:
P = N(E)/N(S) = 240/720 = 1/3;
1 - 1/3 = 2/3.

I seeemed to figure the problem out while typing this hah, oh well that works! I should do this more often.
 
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Yes, that method works nicely. A slightly more concise way to say the ame thing:

There are 6! ways to place the employees at desks. Now, to see how many ways there are of placing them so that the married couple is seated next to one another, treat the married couple as a single "employee". There are 5! ways of placing those 5 "employees", but for each of those, there are 2 ways of ordering the two married people themselves: there are 2(5!) ways of placing them so that the married couple sit beside one another. The probability of that is 2(5!)/6!= 2/6= 1/3. The probability that the married couple are not beside one another, then, is 1- 1/3= 2/3.
 
Thanks for the info Ivy! That does make the answer more clear.
 

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