Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: In how many ways 8 different tickets can be distributed [#permalink]

Show Tags

25 Aug 2015, 20:05

Good exercise. And yet I hope I won't see this kind of ambiguation on the test. Personally, if one person must receive an even number of tickets, the minimum number of tickets is two, not zero. IMO C.

Re: In how many ways 8 different tickets can be distributed [#permalink]

Show Tags

20 Aug 2016, 07:29

I tried to use the approach referred to "Unfair Distributions in Combinatorics - Part 1" but it didnt work for some reason.

So lets say we have 8 DIFFERENT tickets labeled as T1, T2, T3 etc and also we have 2 groups. So imagine put this tickets in a row and also adding any symbol for convince to distinct the two groups. For my example I will use the symbol " + ".

T1 T2 + T3 T4 T5 T6 T7 T8 : This arrangement means Jane gets T1, T2 and Bill gets the rest T3, T4, up to T8.

Another arrangement can be +T1 T2 T3 T4 T5 T6 T7 T8: This arrangement means Jane gets 0 tickets and all of them go to Bill.

Effectively whatever tickets are left hand side from symbol "+" is the "Jane group" and whatever tickets are right hand from " + " is the "Bill group". So classical arrangement problem without repetitions will imply just 9! although the twist occurs because, for instance, the scenario Jane has T1 and T2 is equal to scenario Jane has T2, T1. So there are obvious some duplications and therefore the 9! has to be divided with some magic numbers such that we get the right result. This is where it starts to become complicated and I cannot find these magic numbers.

Re: In how many ways 8 different tickets can be distributed [#permalink]

Show Tags

20 Aug 2016, 08:19

Ndkms wrote:

I tried to use the approach referred to "Unfair Distributions in Combinatorics - Part 1" but it didnt work for some reason.

So lets say we have 8 DIFFERENT tickets labeled as T1, T2, T3 etc and also we have 2 groups. So imagine put this tickets in a row and also adding any symbol for convince to distinct the two groups. For my example I will use the symbol " + ".

T1 T2 + T3 T4 T5 T6 T7 T8 : This arrangement means Jane gets T1, T2 and Bill gets the rest T3, T4, up to T8.

Another arrangement can be +T1 T2 T3 T4 T5 T6 T7 T8: This arrangement means Jane gets 0 tickets and all of them go to Bill.

Effectively whatever tickets are left hand side from symbol "+" is the "Jane group" and whatever tickets are right hand from " + " is the "Bill group". So classical arrangement problem without repetitions will imply just 9! although the twist occurs because, for instance, the scenario Jane has T1 and T2 is equal to scenario Jane has T2, T1. So there are obvious some duplications and therefore the 9! has to be divided with some magic numbers such that we get the right result. This is where it starts to become complicated and I cannot find these magic numbers.

Can anybody explain?

The duplication occur for only 1 case. that is when both receive equal no. of even value. such as 1,2,3,4 for Jane 5,6,7,8 for Bill.

Now when you select 4 out of 8 values, other 4 is already available as a group. So the magic no. here is to divide by 2.

Total no. of groups = 8C4/2 each group can attribute to Jane or bill so 8C4/2*2 = 8c4

for example (ABCD, EFGH) ABCD can be of Jane or Bill so 2 possibility for the group and there are 35 groups

In how many ways 8 different tickets can be distributed [#permalink]

Show Tags

20 Aug 2016, 10:50

abrakadabra21 wrote:

Ndkms wrote:

I tried to use the approach referred to "Unfair Distributions in Combinatorics - Part 1" but it didnt work for some reason.

So lets say we have 8 DIFFERENT tickets labeled as T1, T2, T3 etc and also we have 2 groups. So imagine put this tickets in a row and also adding any symbol for convince to distinct the two groups. For my example I will use the symbol " + ".

T1 T2 + T3 T4 T5 T6 T7 T8 : This arrangement means Jane gets T1, T2 and Bill gets the rest T3, T4, up to T8.

Another arrangement can be +T1 T2 T3 T4 T5 T6 T7 T8: This arrangement means Jane gets 0 tickets and all of them go to Bill.

Effectively whatever tickets are left hand side from symbol "+" is the "Jane group" and whatever tickets are right hand from " + " is the "Bill group". So classical arrangement problem without repetitions will imply just 9! although the twist occurs because, for instance, the scenario Jane has T1 and T2 is equal to scenario Jane has T2, T1. So there are obvious some duplications and therefore the 9! has to be divided with some magic numbers such that we get the right result. This is where it starts to become complicated and I cannot find these magic numbers.

Can anybody explain?

The duplication occur for only 1 case. that is when both receive equal no. of even value. such as 1,2,3,4 for Jane 5,6,7,8 for Bill.

Now when you select 4 out of 8 values, other 4 is already available as a group. So the magic no. here is to divide by 2.

Total no. of groups = 8C4/2 each group can attribute to Jane or bill so 8C4/2*2 = 8c4

for example (ABCD, EFGH) ABCD can be of Jane or Bill so 2 possibility for the group and there are 35 groups

Still though I think 2! is not the full answer + my approach is wrong I think because in my diagram some combinations are not valid. For example

T1 T2 T3 + T4 T5 T6 T7 T8 is not valid sinnce J gets ODD number of tickets. Therefore the "+" can go only to position 1, 3, 5, 7 and 9 so JUST for the "+" there are only 5 ways so 9! is wrong in any case. Still I am missing the full solution according to this method.

Although with my latest correction I think boils down to the solution that mentioned in the sheet. This is because there are 5 ways of having even tickets per person hence individually you have to calculate each case.

Re: In how many ways 8 different tickets can be distributed [#permalink]

Show Tags

29 Aug 2017, 19:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________