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In how many ways 8 different tickets can be distributed

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Joined: 09 Aug 2016
Posts: 72

Kudos [?]: 9 [0], given: 8

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

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New post 20 Aug 2016, 09:50
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.

Kudos [?]: 9 [0], given: 8

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Joined: 22 Jul 2017
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Kudos [?]: 0 [0], given: 23

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

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New post 27 Oct 2017, 15:33
my approach :

Possibilities :
{8,0} , {6,2} , {4,4}, {2,6}, {0,8}

Total no. of ways:
2*(8!/8!)+ 2*8!/(6!*2!) + 8!/(4!*4!)

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

Intern
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Joined: 30 Aug 2017
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Kudos [?]: 2 [0], given: 45

Concentration: Real Estate, Operations
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Re: In how many ways 8 different tickets can be distributed [#permalink]

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New post 28 Oct 2017, 01:28
another solution to this question:

7 tickets can be distributed to 2 people in 2^7 ways ( for each ticket, only 2 choices-either jane or bill. So for 7 tickets, 2x2x2x2x2x2x2=2^7)
Now one person has even# of tickets and other person has odd #. The last ticket(8th) has to be given to the person with odd # of tickets. Thus only 1 choice.

Therefore total ways= 2^7 x 1=2^7=128

Kudos [?]: 2 [0], given: 45

Re: In how many ways 8 different tickets can be distributed   [#permalink] 28 Oct 2017, 01:28

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