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In how many ways 8 different tickets can be distributed
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20 Nov 2009, 07:46
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Re: Sharing tickets.
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22 Nov 2009, 07:25




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20 Nov 2009, 10:26
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.
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Re: Sharing tickets.
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20 Nov 2009, 08:57
Sorry but I do not understand the answer choices. From x to y ?? is it the number of ways?



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20 Nov 2009, 09:04



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22 Nov 2009, 12:52
Nice.... loved this one!
I solved it but it took too much time! Surely I would have jumped



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22 Nov 2009, 16:06
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.
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22 Nov 2009, 16:30
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 (B4 and J4), 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 80 as you can have 08, but in 44 it's only one case. Hope it's clear.
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Re: Sharing tickets.
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23 Nov 2009, 13:23
Bunuel wrote: Everything is right, but the red part.
8C4*4C4 is the formula counting # of ways this can be done (B4 and J4), 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 80 as you can have 08, but in 44 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?
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23 Nov 2009, 16:45
BarneyStinson wrote: Bunuel wrote: Everything is right, but the red part.
8C4*4C4 is the formula counting # of ways this can be done (B4 and J4), 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 80 as you can have 08, but in 44 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 doublecounting.



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23 Nov 2009, 16:56
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 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 doublecounting. aaaarrrrggghhhhhhhh!!!! I get your point!!!! Damn my awesomeness!!!!
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24 Nov 2009, 21:10
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!



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26 Nov 2009, 11:39
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! 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.
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Re: Sharing tickets.
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29 May 2012, 08:26
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?



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08 Oct 2013, 04:31
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^(n1) when n equals to no. of chocolates...therefore 2^(81)=2^7=128.Please correct me if iam wrong.



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Re: In how many ways 8 different tickets can be distributed
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24 Apr 2015, 00:12
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^(n1) when n equals to no. of chocolates...therefore 2^(81)=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
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Re: In how many ways 8 different tickets can be distributed
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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.



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Re: In how many ways 8 different tickets can be distributed
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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.
Can anybody explain?



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Re: In how many ways 8 different tickets can be distributed
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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




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