In how many ways 8 different tickets can be distributed : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 00:01

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In how many ways 8 different tickets can be distributed

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7082

Kudos [?]: 93227 [0], given: 10553

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

### Show Tags

20 Nov 2009, 06:46
Expert's post
8
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

50% (02:48) correct 50% (01:48) wrong based on 262 sessions

### HideShow timer Statistics

In how many ways 8 different tickets can be distributed between Jane and Bill if each is to receive any even number of tickets and all 8 tickets to be distributed.

A. From 2 to 6 inclusive.
B. From 98 to 102 inclusive.
C. From 122 to 126 inclusive.
D. From 128 to 132 inclusive.
E. From 196 to 200 inclusive.
[Reveal] Spoiler: OA

_________________
Director
Joined: 01 Apr 2008
Posts: 897
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 28

Kudos [?]: 645 [0], given: 18

### Show Tags

20 Nov 2009, 07:57
Sorry but I do not understand the answer choices. From x to y ?? is it the number of ways?
Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7082

Kudos [?]: 93227 [0], given: 10553

### Show Tags

20 Nov 2009, 08:04
Economist wrote:
Sorry but I do not understand the answer choices. From x to y ?? is it the number of ways?

Yes. In which range is the correct answer.
_________________
Manager
Joined: 25 Aug 2009
Posts: 175
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 12

Kudos [?]: 176 [9] , given: 3

### Show Tags

20 Nov 2009, 09:26
9
KUDOS
Possibilities

J 8 6 4 2 0

B 0 2 4 6 8

# of ways 8C8 8C6 8C4 8C2 8C0

1+28+70+28+1 = 128 Answer is D.
_________________

Rock On

Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7082

Kudos [?]: 93227 [4] , given: 10553

### Show Tags

22 Nov 2009, 06:25
4
KUDOS
Expert's post
2
This post was
BOOKMARKED
Tickets can be distributed in the following ways:

{8,0} - 8C8=1
{6,2} - 8C6*2C2=28
{4,4} - 8C4*4C4=70
{2,6} - 8C2*6C6=28
{0,8} - 8C8=1

Total # of ways=1+28+70+28+1=128

Answer is D. (Note that this is my question so no OA).

+1 to atish, for the right solution.
_________________
Manager
Joined: 06 Sep 2009
Posts: 115
Followers: 2

Kudos [?]: 25 [0], given: 3

### Show Tags

22 Nov 2009, 11:52
Nice.... loved this one!

I solved it but it took too much time!
Surely I would have jumped
Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 18

Kudos [?]: 164 [2] , given: 22

### Show Tags

22 Nov 2009, 15:06
2
KUDOS
Bunuel wrote:
In how many ways 8 different tickets can be distributed between Jane and Bill if each is to receive any even number of tickets and all 8 tickets to be distributed.

(A) From 2 to 6 inclusive.
(B) From 98 to 102 inclusive.
(C) From 122 to 126 inclusive.
(D) From 128 to 132 inclusive.
(E) From 196 to 200 inclusive.

Again what is wrong with my logic?

I can select 0 tickets for Jane and all 8 tickets for Bill. I could repeat the selection and reverse the order between Bill and Jane.

I can select 2 tickets for Jane and 6 tickets for Bill. I could repeat the selection and reverse the order between Bill and Jane.

I can select 4 tickets for Jane and 4 tickets for Bill. I could repeat the selection and reverse the order between Bill and Jane.

Overall 2[ 8C0 + 8C2 + 8C4 ] = 198.
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7082

Kudos [?]: 93227 [1] , given: 10553

### Show Tags

22 Nov 2009, 15:30
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
SensibleGuy wrote:
Bunuel wrote:
In how many ways 8 different tickets can be distributed between Jane and Bill if each is to receive any even number of tickets and all 8 tickets to be distributed.

(A) From 2 to 6 inclusive.
(B) From 98 to 102 inclusive.
(C) From 122 to 126 inclusive.
(D) From 128 to 132 inclusive.
(E) From 196 to 200 inclusive.

Again what is wrong with my logic?

I can select 0 tickets for Jane and all 8 tickets for Bill. I could repeat the selection and reverse the order between Bill and Jane.

I can select 2 tickets for Jane and 6 tickets for Bill. I could repeat the selection and reverse the order between Bill and Jane.

I can select 4 tickets for Jane and 4 tickets for Bill. I could repeat the selection and reverse the order between Bill and Jane.

Overall 2[ 8C0 + 8C2 + 8C4 ] = 198.

Everything is right, but the red part.

8C4*4C4 is the formula counting # of ways this can be done (B-4 and J-4), we don't need to multiply this by 2.

You don't need to reverse the order in this case as ALL possible scenarios are already covered with 8C4.

You are reversing in scenario 8-0 as you can have 0-8, but in 4-4 it's only one case.

Hope it's clear.
_________________
Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 18

Kudos [?]: 164 [1] , given: 22

### Show Tags

23 Nov 2009, 12:23
1
KUDOS
Bunuel wrote:
Everything is right, but the red part.

8C4*4C4 is the formula counting # of ways this can be done (B-4 and J-4), we don't need to multiply this by 2.

You don't need to reverse the order in this case as ALL possible scenarios are already covered with 8C4.

You are reversing in scenario 8-0 as you can have 0-8, but in 4-4 it's only one case.

Hope it's clear.

Let me be more clear.

I can select 4 tickets of 8 in 8C4 ways. Remaining 4 tickets will be 4C4 obviously.
After selecting those 4 tickets, I have to select either Bill or Jane as the receiver of the 4 tickets in 2C1 ways. The remaining person is 1C1.
After making the two selections, all possible combinations are 8C4*4C4*2C1*1C1 = 70*2 = 140.

Is there a flaw in my selections?
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 5

Kudos [?]: 341 [0], given: 6

### Show Tags

23 Nov 2009, 15:45
BarneyStinson wrote:
Bunuel wrote:
Everything is right, but the red part.

8C4*4C4 is the formula counting # of ways this can be done (B-4 and J-4), we don't need to multiply this by 2.

You don't need to reverse the order in this case as ALL possible scenarios are already covered with 8C4.

You are reversing in scenario 8-0 as you can have 0-8, but in 4-4 it's only one case.

Hope it's clear.

Let me be more clear.

I can select 4 tickets of 8 in 8C4 ways. Remaining 4 tickets will be 4C4 obviously.
After selecting those 4 tickets, I have to select either Bill or Jane as the receiver of the 4 tickets in 2C1 ways. The remaining person is 1C1.
After making the two selections, all possible combinations are 8C4*4C4*2C1*1C1 = 70*2 = 140.

Is there a flaw in my selections?

By using 8C4 exclusively, you are already counting the possibility of a specific set of 4 tickets going to either person.

For example, say you had 4 tickets, labeled A, B, C and D.

4C2 = 6
AB, CD
AC, BD

As you can see, it accounts for a specific set of 2 tickets, as well as the complete opposite. By multiplying by a factor of 2, you end up double-counting.
Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 18

Kudos [?]: 164 [0], given: 22

### Show Tags

23 Nov 2009, 15:56
AKProdigy87 wrote:
By using 8C4 exclusively, you are already counting the possibility of a specific set of 4 tickets going to either person.
For example, say you had 4 tickets, labeled A, B, C and D.

4C2 = 6
AB, CD
AC, BD

As you can see, it accounts for a specific set of 2 tickets, as well as the complete opposite. By multiplying by a factor of 2, you end up double-counting.

aaaarrrrggghhhhhhhh!!!! I get your point!!!! Damn my awesomeness!!!!
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Manager
Joined: 18 Nov 2009
Posts: 57
Followers: 0

Kudos [?]: 9 [0], given: 7

### Show Tags

24 Nov 2009, 20:10
hello, so in the end, the answer is 128 right? i got the same answer, using the same solution as atish. i'm just making sure it's the right one.

i'm also wondering as to the answer choices: why the range? shouldn't there be one and only one answer? is there a way to approximate the answer? anyone have any ideas as to approximating this? i solved the problem, but it took me about 4 minutes, which is time i wouldn't have in the actual exam.

cheers!
Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7082

Kudos [?]: 93227 [0], given: 10553

### Show Tags

26 Nov 2009, 10:39
pochcc wrote:
hello, so in the end, the answer is 128 right? i got the same answer, using the same solution as atish. i'm just making sure it's the right one.

i'm also wondering as to the answer choices: why the range? shouldn't there be one and only one answer? is there a way to approximate the answer? anyone have any ideas as to approximating this? i solved the problem, but it took me about 4 minutes, which is time i wouldn't have in the actual exam.

cheers!

Answers are represented as ranges, because there are number of incorrect answers possible and several ranges are needed to cover them all.
_________________
Senior Manager
Joined: 28 Dec 2010
Posts: 334
Location: India
Followers: 1

Kudos [?]: 201 [0], given: 33

### Show Tags

29 May 2012, 07:26
Hey bunuel the question steam says tickets are to distributed such that both get even no. correct me, isnt 0 neither even nor odd? then how can we take the case of 8,0?
Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7082

Kudos [?]: 93227 [0], given: 10553

### Show Tags

29 May 2012, 08:00
vibhav wrote:
Hey bunuel the question steam says tickets are to distributed such that both get even no. correct me, isnt 0 neither even nor odd? then how can we take the case of 8,0?

Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13459
Followers: 575

Kudos [?]: 163 [0], given: 0

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

### Show Tags

06 Oct 2013, 11:26
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.
_________________
Intern
Joined: 21 Sep 2013
Posts: 30
Location: United States
Concentration: Finance, General Management
GMAT Date: 10-25-2013
GPA: 3
WE: Operations (Mutual Funds and Brokerage)
Followers: 0

Kudos [?]: 23 [0], given: 82

### Show Tags

08 Oct 2013, 03:31
Bunuel wrote:
vibhav wrote:
Hey bunuel the question steam says tickets are to distributed such that both get even no. correct me, isnt 0 neither even nor odd? then how can we take the case of 8,0?

Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

Hope it's clear.

I dunno my approach is right or wrong but i get the same ans.

by applying the formula 2^(n-1) when n equals to no. of chocolates...therefore 2^(8-1)=2^7=128.
Please correct me if iam wrong.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13459
Followers: 575

Kudos [?]: 163 [0], given: 0

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

### Show Tags

06 Nov 2014, 10:06
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.
_________________
Intern
Joined: 10 Jan 2015
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 9

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

### Show Tags

23 Apr 2015, 19:12
Bunuel wrote:
vibhav wrote:
Hey bunuel the question steam says tickets are to distributed such that both get even no. correct me, isnt 0 neither even nor odd? then how can we take the case of 8,0?

Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

Hope it's clear.

Bunuel by that logic even -2 should also be even as it is of a form 2k where k is -1
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13659 [1] , given: 222

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

### Show Tags

23 Apr 2015, 23:12
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Yash12345 wrote:
Bunuel wrote:
vibhav wrote:
Hey bunuel the question steam says tickets are to distributed such that both get even no. correct me, isnt 0 neither even nor odd? then how can we take the case of 8,0?

Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

Hope it's clear.

I dunno my approach is right or wrong but i get the same ans.

by applying the formula 2^(n-1) when n equals to no. of chocolates...therefore 2^(8-1)=2^7=128.
Please correct me if iam wrong.

Your approach is very good and the reason it works is this:

You have 8 different tickets. Each ticket can be given to one of the 2 people in 2 ways. You do that with 7 tickets in 2^7 ways. When you distribute 7 tickets, one person will have odd number of tickets and one will have even number of tickets (0 + 7 or 1 + 6 or 2 + 5 or 3 + 4).
The eighth ticket needs to be given to the person who has odd number of tickets so you give the 8th ticket in only one way. This accounts for all cases in which both get even number of tickets.

Total ways = 2^7 * 1 = 128
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Re: In how many ways 8 different tickets can be distributed   [#permalink] 23 Apr 2015, 23:12

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
4 In how many ways can 5 different fruits be distributed among 4 26 Jun 2014, 22:53
41 In how many ways can 5 different marbles be distributed in 4 19 27 Apr 2014, 23:59
14 In how many ways can 5 different candies be distributed in 15 21 Oct 2012, 17:00
15 In how many ways can 5 different candiesbe distributed among 13 21 Oct 2012, 16:52
18 In how many different ways can a group of 8 people be 11 12 Jan 2012, 06:59
Display posts from previous: Sort by