Probabilty: 2 couples and 1 single person.

[nayak.purnendu]

Source: PR Test 1

Q> Two couples and one 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


Can someone explain me how to solve this ?
OA: (d)

[msgrewal81]

I was gonna ask similar question but lets work on this one. This is a way I found on MGMAT but I know I am wrong. I have test on Saturday, please somebdy guide where my strategy is wrong. I am late but prob. getting the heck out of me

Total Possible ways to seat = 5!

Now, assume 1st couple, 2nd couple, 1 guy. Now the # of ways they can arrange = 3!, but since we have to account for couples exchanging seats bw themselves so the actual # of arrangements = 3! x 2! x 2!

Therefore, probabilty that couples will be seated together = (3! x 2 x 2) / 5! = 1/5

Therefore, probability that couple NOR seated together = 1 - 1/5 = 4/5, which is wrong answer and I invite somebdy to correct my approach.

[ilakshmir]

Here is how I would approach this problem:

Whenever there is a requirement on coupling the position of 2 or more entities, use the following trick:

If out of 5 people, 2 must sit together, in how many ways can they be arranged?
Let the # of ways this achieved by K. Let us couple the 2 people (treat them as a single entity). So there are only 4 entities. Now permute them --> 4! ways exist. Within the coupled entity, the 2 people can be arranged in 2 ways, (AB, BA). So there now exists 2x4! ways of arranging 5 people if 2 have to sit next to each other. So K= 2x4!.

Now let us attack the given problem:
- 5 people can be arranged in general in 5! ways.
- we have 2 couples that must be separated. We will begin with combinations where the couple is together and subtract this from 5!.
- The # of ways a couple can be together: The 1st couple will be separated in K = 2x4! ways. The 2nd couple will be separated in K = 2x4! ways. Ways by which 1st and 2nd couple will be separated = 2x4!x2 - (common arrangements)
- Common arrangements C occur when both couples are together. This will occur in 3! ways (using logic in 1st paragraph). Each couple can be seated in 2 ways when they are treated as a single entity. So C = 2x2x3!

So the final solution is (5! - 2x4! - 2x4! + 2x2x3!)/5! = 2/5

[shaji]

Good effort Ilakshmir!!!. Perfect logic in the first two paragraphs of your explaination excepting the final solution though the final outcom is correct. Some would not hesitate to say you got 'lucky', but perhaps others might comment you made a 'compensating error' while yet some others would call it a 'good mistake'. Needless to mention, the 'good mistake' might have been provoked by hindsight or in other words, an effort of working towards the correct answer. Neverthess the 'good mistake' is indeed very stimulating.
Now the correct approach.
The number of ways that none of the families sit together is:=(Total number of ways of seating-(number of ways just one family is seated together+both families are seated together.)
Which is numerically:-
(5!-(48+24))=48
The required probability is therefore=48/120=2/5.
Try the same problem if there are three families and one single to appreciate the logic.

[ilakshmir]

shaji

When I provided my solution, I did not arrive at it through serendipity nor did I look for a way to somehow match the given answer. I still stand by my solution.

In your approach, the number of ways just one family is seated together = 48,
both families are seated together = 24

These are indeed the values I have derived in my solution. What is it that you disagree with?

I request an MGMAT instructor to moderate...

[shaji]

Ilakshmir;
"These are indeed the values I have derived in my solution" . They fact of the matter is that it isn't evident that the 'values' are derived since they can't be derived.
Try the same problem if there are three families and one single to understand the error in your logic. You get into the trap of multiple doule counting which you were lucky to get away with the two families problem.

[ilakshmir]

Shaji

how do you compute the number of ways just one family is seated together = 48? Maybe that will help me understand why you say I have not derived this number.

[shaji]

Ilakshmir;
"how do you compute the number of ways just one family is seated together = 48?"
Here it goes:-
The number of ways that lucky family gets to sit together is 2*2=4 ways, and for each of those ways the unlucky family can be seperated by 1)both the family as well as single(2*2=4ways) or 2)by either the family or the single person(2*2*2=8 ways)
The total number of ways therefore=(4*(8+4))=48

[jigar24]

Hi, I have a slightly different approach -

Now lets call the couples 'a' 'a1' and 'b' 'b1' and the single person as 'c'

The position of 'c' will determine the permutations

1) If 'c' occupies the first seat:

C _ _ _ _

The second seat can be filled in 4 ways (a, a1, b, b1)
The third seat can bee filled in 2 ways (for eg. if 'a' filled the second seat, 'b' or 'b1' could be seated in the 3rd)
The fourth seat can be filled in 1 way (a1)
The fifth seat can be filled in 1 way (b1)

This way none of the couples sit next to each other.

Thus total permutations = 1 * 4 * 2 * 1 * 1 = 8ways

Now apply the same logic to the remaining 4 cases

2) If 'c' occupies second seat -

_ c _ _ _

4 * 1 * 2 * 1 * 1 = 8 ways

3) If 'c' occupies third seat -

_ _ c _ _

4 * 2 * 1 * 2 * 1 = 16 ways

4) If 'c' occupies fourth seat -

_ _ _ c _

4 * 2 * 1 * 1 * 1 = 8 ways

5) If 'c' occupies fifth seat -

_ _ _ _ c

4 * 2 * 1 * 1 * 1 = 8 ways

SO total ways in which they can be seated without any couple sitting next to each other = 8 + 8 + 16 + 8 + 8 = 48

Total ways in which they can be arranged = 5!

Thus the required probability = 48/120 = 2/5

[shaji]

jigar24
Perfect logical approach!!!. Will do for the two family problem but as the number of families increases and in particular odd numbers, matters may get cumbersome from the viewpoint of time.

[RonPurewal]

I was gonna ask similar question but lets work on this one. This is a way I found on MGMAT but I know I am wrong. I have test on Saturday, please somebdy guide where my strategy is wrong. I am late but prob. getting the heck out of me

Total Possible ways to seat = 5!

Now, assume 1st couple, 2nd couple, 1 guy. Now the # of ways they can arrange = 3!, but since we have to account for couples exchanging seats bw themselves so the actual # of arrangements = 3! x 2! x 2!

Therefore, probabilty that couples will be seated together = (3! x 2 x 2) / 5! = 1/5

Therefore, probability that couple NOR seated together = 1 - 1/5 = 4/5, which is wrong answer and I invite somebdy to correct my approach.

you are using false logic. you are assuming that the following two events are opposites:
* BOTH couples are seated together (this is the event whose probability you have just found)
* NEITHER couple is seated together (the event that is the subject of the question)

problem is, these aren't opposite events. if you're going to approach it this way - find the opposite probability and then subtract - you'll have to find the probability you just found, AND find the probability that exactly ONE of the couples is seated together. then you'll have to add those together, and subtract the SUM from 1.

hope that helps