Let's find the opposite probability and subtract it from 1.

Opposite event that neither of the couples sits together is event that at leas one couple sits together. # of arrangements when at leas one couple sits together is sum of arrangements when EXACTLY 2 couples sit together and EXACTLY 1 couples sit together.

Couple A: A1, A2
Couple B: B1, B2
Single person: S

EXACTLY 2 couples sit together:
Consider each couple as one unit: {A1A2}{B1B2}{S}, # of arrangement would be: 3!*2!*2!=24. 3! # of different arrangement of these 3 units, 2! arrangement of couple A (A1A2 or A2A1), 2! arrangement of couple B (B1B2 or B2B1).

EXACTLY 1 couple sits together:
Couple A sits together: {A1A2}{B1}{B2}{S}, # of arrangement would be: 4!*2!=48. 4! # of different arrangement of these 4 units, 2! arrangement of couple A (A1A2 or A2A1). But these 48 arrangements will also include arrangements when 2 couples sit together, so total for couple A would be 48-24=24;

The same for couple B: {B1B2}{A1}{A2}{S}, # of arrangement would be: 4!*2!=48. Again these 48 arrangements will also include arrangements when 2 couples sit together, so total for couple B would be 48-24=24;

24+24=48.

Finally we get the # of arrangements when at least one couple sits together is 24+48=72.

Total # of arrangements of 5 people is 5!=120, hence probability of an event that at leas one couple sits together would be \frac{72}{120}=\frac{3}{5}.

So probability of an event that neither of the couples sits together would be 1-\frac{3}{5}=\frac{2}{5}

Answer: D.

Hope it's clear.
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I think Bunuel's expanation is correct. This question is about arrangements so Combination will have to be refered to solve it.
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Hi would someone please help me with this one:

Two couples and on single person are seated at random in a row of five chairs. what is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2


I seem to be able to come up with the right numbers but I can't figure why I would put them together.

5 seats so, 5! options for people to be arranged = 120. And If the one of the couples sit together then there are 4! ways for the others to sit (24), since there are 2 couples then 2x4! (48).

But even if this is right I don't know why I would end up setting the answer so that 4!+4!/5! (48/120) as the probability. How can one convert the different combinations of something into the probability that something happens?

Maybe its just coincidence that that ends up being the right answer.

thanks

Can you pls explain again..
I was not able to understand the second statement of the explanation...

Thanks to heyholetsgo. I think that P(A or B) = P(A) + P(B) - P(A and B) is the best approach:

Combinations for one couple sitting next to each other 4! x 2! ( couple arrange in 2! ways between themselves)=24x2=48
Combinations for the other couple sitting next to each other 4! x 2! ( couple arrange in 2! ways between themselves)=24x2=48

To calculate P(A and B) we glue both couples together
3! x 2! x 2! ( each couple arrange in 2! ways between themselves)
= 24

P(A) + P(B) - P(A and B)
48 + 48 -24=72
1-(72/120)=2/5

Total ways to arrange = 5! = 120
Ways in which 1st couple can sit together = 4! X 2! = 48
Ways in which both couple sit together = 3! X 2! X 2! = 24
Ways in which ONLY 1st couple can sit together = 48-24 = 24 = Ways in which ONLY 2nd couple sit together
Ways in which neither couple sits together = Total ways – Ways in which both sit together –Ways in which only 1st couple sits together – Ways in which 2nd couple sits together
= 120-24 –24-24 = 120-72 = 48
Probability = 48/120 = 2/5

I think the best combinatorial solution was given by heyholetsgo. Neat and short!

I didn't see a probabilistic approach, so I tried to work out one. Here it is:

There are 5 chairs, let's visualize them _ _ _ _ _
There is one single person, let's denote him/her by S.
S can choose from 5 chairs where to sit. There are some symmetrical placements for S:
1) S _ _ _ _ and _ _ _ _ S (0, 1, or 2 couples can be placed together)
2) _ S _ _ _ and _ _ _S_ (0 or 1 couple can be placed together)
And there is the third case:
3) _ _ S _ _ (0, 1, or 2 couples can be placed together).

So, let's compute for each case the probability that 0 couple sits together.

1) S _ _ _ _ or _ _ _ _ S:
\frac{2}{5} probability that S sits on either chair 1 or chair 5.
Now start placing people on the remaining four chairs and express the corresponding probabilities:
\frac{4}{4} for the first chair in the sequence of four remaining chairs
\frac{2}{3} for the second chair (cannot be the first person's mate)
\frac{1}{2} for the third chair (cannot be the mate of the second person)
\frac{1}{1} for the last person to be placed
In conclusion, the probability of no pairs sittings together when S sits on either chair 1 or 5 is given by
\frac{2}{5}*\frac{4}{4}*\frac{2}{3}*\frac{1}{2}*\frac{1}{1}=\frac{2}{15}.

2) _ S _ _ _ or _ _ _ S _
Again, two possibilities for S, so \frac{2}{5} probability for S to sit on either chair 2 or 4.
Now start placing people on the remaining four chairs and express the corresponding probabilities.
Treat only the _ S _ _ _ case, the other one has the same probabilities:
\frac{4}{4} for the first chair from the left
\frac{2}{3} for the second chair (cannot be the first person's mate because then the last two chairs will be occupied by the other couple)
\frac{1}{2} for the third chair (cannot be the mate of the second person)
\frac{1}{1} for the last person to be placed
In conclusion, the probability of no pairs sittings together when S sits on either chair 2 or 4 is given by
\frac{2}{5}*\frac{4}{4}*\frac{2}{3}*\frac{1}{2}*\frac{1}{1}=\frac{2}{15}.

3) _ _ S _ _
Here now we just have one possibility to place S, so \frac{1}{5} probability for S to sit on chair 3.
Starting from the left, place people on chairs:
\frac{4}{4} for the first chair from the left
\frac{2}{3} for the second chair (cannot be the first person's mate)
\frac{2}{2} for the third chair (we are already assured that we separated the two couples)
\frac{1}{1} for the last person to be placed
In conclusion, the probability of no pairs sittings together when S sits on chair 3 is given by
\frac{1}{5}*\frac{4}{4}*\frac{2}{3}*\frac{2}{2}*\frac{1}{1}=\frac{2}{15}.

In conclusion, the requested probability is 3*\frac{2}{15}=\frac{2}{5}.

Answer D
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Instead of doing so many calculations, why not straight up find the probability that no one sits together?

Sure, you can do the venn diagram approach and find individual circles and subtract the overlap, but this way was a bit easier for me to understand:

_ _ _ _ _ = 5 positions

The single person can sit in 3 spots (Later we will account for all the permutations by multiply by 3! = 6)
S _ _ _ _ (sitting at the end)
_ S _ _ _ (Sitting 1 spot from the end)
_ _ S _ _ (sitting in the middle)

Note the remaining options (_ _ _ S _ and _ _ _ _ S) are just inverses of the first two.

So let's look at the first one: S _ _ _ _

S A1 B1 A2 B2
S A1 B2 A2 B1
S B1...
S B2...

If we pick A1 for that 2nd spot, then the 3rd spot cannot be A2 since A's cannot since next to each other. That 3rd spot must be one of the 2 B's: B1 or B2. Then the 4th spot must be the remaining A2. Then the last spot will be the remaining B.

Note that A and B must alternate - NO MATTER where S is located.

This is true for the other 2 forms (_ S _ _ _) as well as (_ _ S _ _ ) - A's and B's must alternate.

So back to S _ _ _ _

That second spot S (_) _ _ _ has 4 options to choose from (A1, A2, B1, and B2). Whichever you choose for that second spot, the third spot you will have 2 options to choose from.

If you choose A1 for the 2nd spot, then the 3rd spot has 2 options (B1 or B2). The remaining 4th and 5th spots are self-chosen so there is no need to calculate those last spots in this case.

S A1 B1 A2 B2
S A1 B2 A2 B1

So here's the calculation:

(Out of 4 options A1, A2, B1, and B2- choose 1) * (Out of remaining 2 options, choose 1)

= (4C1) * (2C1) = 8

Multiply by 3! for each of the positions that "S" can occupy and all the permutations around it (including S _ _ _ _ and _ _ _ _ S, etc)

So we have 8 * 3! = 8 * 6 = 48 possibilities that no couples sit together

Since there are a total of 5 positions, the total possibilities is 5! = 120
You can also think of this as (5C1 * 4C1 * 3C1 * 2C1 * 1C1) = 5! = 120

So divide the 48 possibilities that no couples sit together by the total 120 possibilities and you get:

48/120 = 24/60 = 4 / 10 = 40%
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