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Alternate Method To Solve This Counting Problem

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Alternate Method To Solve This Counting Problem

by knight247 » Fri Sep 16, 2011 2:51 am
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A)5%
(B)10%
(C)15%
(D)20%
(E)25%

Yeah I know the answer is B. We solve considering Bob and Lisa sitting at the extreme ends. So we have Bob X Y Z Lisa which can be arranged in 6 ways or Lisa X Y Z Bob which can be arranged in 6 ways. So we have a total of 12 possible outcomes. The total number of possible outcomes is 5!=120. So Prob=12*100/120=10% hence B

Now with the intention of strengthening my basics in counting, hoping someone can help me with an alternate solution to this one. Maybe considering all the scenarios where Bob and Lisa will only be in the middle 3 seats. Which is subtracted from the total possibilities with no restrictions etc. Alternate approaches to this problems would be appreciated. Thanks
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by saketk » Fri Sep 16, 2011 3:07 am
knight247 wrote:A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A)5%
(B)10%
(C)15%
(D)20%
(E)25%

Yeah I know the answer is B. We solve considering Bob and Lisa sitting at the extreme ends. So we have Bob X Y Z Lisa which can be arranged in 6 ways or Lisa X Y Z Bob which can be arranged in 6 ways. So we have a total of 12 possible outcomes. The total number of possible outcomes is 5!=120. So Prob=12*100/120=10% hence B

Now with the intention of strengthening my basics in counting, hoping someone can help me with an alternate solution to this one. Maybe considering all the scenarios where Bob and Lisa will only be in the middle 3 seats. Which is subtracted from the total possibilities with no restrictions etc. Alternate approaches to this problems would be appreciated. Thanks

This is how I solved..

apart from Bob and Lisa there are 3 people.

one out of 3 can be selected in 3 ways - bob & lisa will sit next to this person all the time. Let's name him A

This also means that these 3 will always sit together. let's make them 1 group. let's call this group G

then we have 3 people to arrange (a group and 2 people[M,N])

M,N & G.

These 3 can be arranged in 6 ways .. in G we will have 2 possible combo -- Bob A Lisa or Lisa A Bob

= 6*2= total 12 ways

12/120 = 1/10
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by GMATGuruNY » Fri Sep 16, 2011 3:24 am
knight247 wrote:A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A)5%
(B)10%
(C)15%
(D)20%
(E)25%

Yeah I know the answer is B. We solve considering Bob and Lisa sitting at the extreme ends. So we have Bob X Y Z Lisa which can be arranged in 6 ways or Lisa X Y Z Bob which can be arranged in 6 ways. So we have a total of 12 possible outcomes. The total number of possible outcomes is 5!=120. So Prob=12*100/120=10% hence B

Now with the intention of strengthening my basics in counting, hoping someone can help me with an alternate solution to this one. Maybe considering all the scenarios where Bob and Lisa will only be in the middle 3 seats. Which is subtracted from the total possibilities with no restrictions etc. Alternate approaches to this problems would be appreciated. Thanks
Bob and Lisa must each sit next to exactly one of the 4 other people.
Thus, Bob and Lisa must sit at the ends.

Good arrangements:
Number of options for the leftmost seat = 2. (Bob or Lisa.)
Number of options for the rightmost seat = 1. (Either Bob or Lisa, whoever is not sitting in the leftmost seat.)
Number of ways to arrange the 3 remaining seats = 3*2*1.
To combine these options, we multiply:
Good arrangements = 2*1*3*2*1.

Total possible arrangements = 5*4*3*2*1.

Good/Total = (2*1*3*2)/(5*4*3*2) = 1/10 = 10%.

The correct answer is B.
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by bijoyajj » Fri Sep 16, 2011 7:28 am
Hi GMATGuruNY,

If the question is like below then how can we use your same approach?

"what is the probability that Bob and Lisa will each sit next to only one of the three other students from the group? "

So in addition they can sit in 2nd & 3rd OR 3rd and 4th seats with other 3 occupying the seat next to them.. so is it 16/120 now?
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by ashutoshkumar7 » Fri Sep 16, 2011 7:42 am
What is the answer if Bob and List are not allowed to take the corner seats?
Thanks,
Ashutosh
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by GMATGuruNY » Fri Sep 16, 2011 8:02 am
bijoyajj wrote:Hi GMATGuruNY,

If the question is like below then how can we use your same approach?

"what is the probability that Bob and Lisa will each sit next to only one of the three other students from the group? "

So in addition they can sit in 2nd & 3rd OR 3rd and 4th seats with other 3 occupying the seat next to them.. so is it 16/120 now?
Case 1: Bob and Lisa occupy the two seats on the end.
Number of options for the leftmost seat = 2. (Bob or Lisa.)
Number of options for the rightmost seat = 1. (Either Bob or Lisa, whoever is not sitting in the leftmost seat.)
Number of ways to arrange the 3 remaining seats = 3*2*1.
To combine these options, we multiply:
2*1*3*2*1 = 24.

Case 2: Bob and Lisa sit next to each other and occupy two middle seats.
Number of options for the leftmost seat = 3. (Anyone but Bob and Lisa.)
Number of options for the rightmost seat = 2. (Anyone but Bob, Lisa, and the person in the leftmost seat.)
Number of options for the middle seat = 2. (Must be Bob or Lisa.)
Number of ways to arrange the two remaining seats = 2*1.
To combine these options, we multiply:
3*2*2*2*1 = 24.

Total good arrangements = 24+24 = 48.
Total possible arrangements = 5*4*3*2*1 = 120.

(Total good arrangements)/(Total possible arrangements) = 48/120 = 2/5.
Last edited by GMATGuruNY on Fri Sep 16, 2011 8:37 am, edited 3 times in total.
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by bijoyajj » Fri Sep 16, 2011 8:12 am
ashutoshkumar7 wrote:What is the answer if Bob and List are not allowed to take the corner seats?
Going by his method we will get it like 3*2* 3!

first seat can be occupied in 3 ways (since L&b can't occupy side seats)
Last seat can be occupied in 2 ways

now 3 people occupying 3 seats in the middle = 6 ways

so total 36 ways i guess..
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by bijoyajj » Fri Sep 16, 2011 8:18 am
GMATGuruNY wrote:
In order for Bob and Lisa each to sit next to exactly one of the THREE other people, Bob and Lisa must sit next to each other and occupy two middle seats.
Hi GMATGuruNY,

L & B occupying the side seats too fall into the above scenario.Isn't that?.. in that case also only other 3 sits near them.. so then wat would be the solution?
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by GMATGuruNY » Fri Sep 16, 2011 8:36 am
bijoyajj wrote:Hi GMATGuruNY,

L & B occupying the side seats too fall into the above scenario.Isn't that?.. in that case also only other 3 sits near them.. so then wat would be the solution?
Yes, indeed. Please see my amended post above.
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by Brent@GMATPrepNow » Fri Sep 16, 2011 10:28 am
knight247 wrote:A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A)5%
(B)10%
(C)15%
(D)20%
(E)25%

Yeah I know the answer is B. We solve considering Bob and Lisa sitting at the extreme ends. So we have Bob X Y Z Lisa which can be arranged in 6 ways or Lisa X Y Z Bob which can be arranged in 6 ways. So we have a total of 12 possible outcomes. The total number of possible outcomes is 5!=120. So Prob=12*100/120=10% hence B
This need not be a counting question. You have already identified that we need Bob and Lisa to sit on the extreme ends. So, we can reword the question as "What is the probability that Bob and Lisa both sit on the extreme ends?"

P(both sit on extreme ends) = P(one of them sits on left-most seat AND the other sits on the right-most seat)

= P(one of them sits on left-most seat) X P(the other sits on the right-most seat)

= (2/5) X (1/4)
= 1/10
= B

Cheers,
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