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

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20 Nov 2009, 07:46

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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.

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.

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.

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? _________________

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 AD, BC

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.

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 AD, BC

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!!!! _________________

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.

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!

Yes, the answer is 128.

Answers are represented as ranges, because there are number of incorrect answers possible and several ranges are needed to cover them all. _________________

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?

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 \(k=0\) --> \(n=2*0=0\).

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

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06 Oct 2013, 12:26

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).

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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.

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

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06 Nov 2014, 11:06

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. _________________

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

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23 Apr 2015, 20:12

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

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

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24 Apr 2015, 00:12

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Expert's post

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.

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