probability

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sharmishtha_goel: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Thu Feb 17, 2011 9:13 am; Thanked: 1 times

by sharmishtha_goel » Tue Aug 02, 2011 12:46 pm

One single person and two couples are to be 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

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Source: Beat The GMAT — Problem Solving | Tue Aug 02, 2011 12:46 pm

rishimaharaj: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Mon May 02, 2011 11:18 amFollowed by: 7 members; Location: Florida; Thanked: 20 times; GMAT Score:710

by rishimaharaj » Tue Aug 02, 2011 6:46 pm

Hello Sharmishtha,

In the Manhattan GMAT math workbooks, they have the idea of creating anagrams with the problem to help with counting the number of ways.

If you want a total count, you just take the factorial.
If you have repeats, or things "stuck together" take the factorial, divided by each "stuck together" factorial.

In our problem, we will find the following:
-Total number of ways they can be placed. = P(Total)
-Total number of ways they can be placed WHERE Couple-1 is together. = P(C1)
-Total number of ways they can be placed WHERE Couple-2 is together. = P(C2)
-Total number of ways they can be placed WHERE BOTH Couple-1 and Couple-2 are together. = P(Both) = P(C1 âˆ© C2)

P(Total) = A B C D E
= 5*4*3*2*1
= 5!
= 120
-There are five people, after placing the first, there are four people, etc.. Multiplication Rule applies, and we get 5! possible ways to arrange them.

P(C1) = A A B C D
= (5!) / (2!)
= 120 / 2
= 60
Couple 1 sits together, so they are "stuck together" thus we divide by the factorial of the number of A's in our anagram (2 A's, so 2!). There are 60 ways to arrange the 5 people where couple 1 is sitting next to each other.

P(C2) = A B C C D
= (5!) / (2!)
= 120 / 2
= 60
Same as above. Couple 2 sits together, so there are 2 C's, so 60 possible ways to arrange them with that seating.

P(C1 âˆ© C2) = A A B B C
= (5!) / (2!)(2!)
= 120 / 4
= 30
Both Couple 1 and 2 are sitting next to each other, hence we divide by 2! twice. 30 ways to arrange the 5 people and still have both couples sitting next to their respective partners.

The formula for Union is P(C1 U C2) = P(C1) + P(C2) - P(C1 âˆ© C2)
P(C1 U C2) = 60 + 60 - 30
= 120 - 30
= 90
So there are 90 ways to arrange these 5 people where: C1 is together, C2 is together, or both C1 and C2 are together.

SO, out of the 120 ways, 90 ways they are together. So they ways they are not together is:
120 - 90 = 30.

The probability of neither of them sitting together is
= (# of ways they are not together) / (total # of arrangements)
= 30 / 120
= 3/12
= 1/4

The answer is B.

Hope this helps! If you have any doubts, please let us know!

--Rishi

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gmatboost: Master | Next Rank: 500 Posts; Posts: 312; Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members; Location: New York City; Thanked: 130 times; GMAT Score:780

by gmatboost » Tue Aug 02, 2011 7:34 pm

This is a very hard question in my opinion. Unfortunately, though Rishi hit on a lot of good points, that answer is not correct.

In your P(C1) and P(C2) calculations 60 is not actually the number of ways that AA can sit together. The formula 5!/2! tells us the number of unique ways we can arrange 5 elements if 2 of them are considered to be indistinguishable. In other words, A1,A2,B,C,D and A2,A1,B,C,D would be considered just 1 arrangement: AABCD. So, 60 is the total number of ways that we can arrange AABCD.

To find the true values of P(C1) and P(C2):
There are 4 ways in which AA can sit together: AAxxx, xAAxx, xxAAx, and xxxAA. In each case, the two As can sit A1,A2 or A2,A1 so there are 4*2 = 8 ways they can sit together.
In each of 8 seatings, the other three can be arranged in 3! = 6 ways, so P(C1) = P(C2) = 8*6 = 48.

You are right that we should subtract the overlap: the number of ways they can both sit together. However, there are also not 30 ways this can happen.

For both couples to sit together, C must be in seat 1, 3, or 5. If C is in 2 or 4, it's not possible.
In each of those three cases, AA and BB can fill the other 4 slots, from left to right, either as AABB or BBAA. So there are 2 options there. Within each option, AA can sit in 2 ways and BB can sit in two ways, so there are 2*2*2 = 8 ways for them to sit in each of the 3 locations for C. So, there are 3*8 = 24 ways for them to both sit together.

Then, since 48 + 48 - 24 = 72, there are 72 ways for them to sit together, and 48 ways for them not to. 48/120 = 2/5.

A totally different approach.

Case 1: C is in spot 1 or 5 (these situations are reflections of each other and therefore identical). Probability = 2/5. Put A1 in the spot next to C. So: C,A1,x,y,z.
If A2 goes in spot y, both couples will NOT be together, otherwise at least one will. So, P(not together) in Case 1 = 1/3.

Case 2: C is in spot 2 or 4 (these situations are reflections of each other and therefore identical). Probability = 2/5. Put A1 in the spot outside of C. So: A1,C,x,y,z.
Again, if A2 goes in spot y, both couples will NOT be together, otherwise at least one will. So, P(not together) in Case 2 = 1/3.

Case 3: C is in spot 3. Probability = 1/5. Put A1 in spot 2. So: x,A1,C,y,z.
Here, if A2 goes in spot y OR z, both couples will NOT be together, otherwise both will. So, P(not together) in Case 3 = 2/3.

Total P(not together) = 2/5 * 1/3 + 2/5 * 1/3 + 1/5 * 2/3 = 2/15 + 2/15 + 2/15 = 6/15 = 2/5.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Aug 03, 2011 2:18 am

I posted a solution here:

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