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In how many ways 8 different tickets can be distributed

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Ndkms

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20 Aug 2016, 10:50

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abrakadabra21

Ndkms

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.

radhikakhemka007

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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!)

Ace800

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28 Oct 2017, 02:28

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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

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15 Jan 2018, 11:51

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Bunuel

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.

hi

I have seen a solution to a problem similar to this one elsewhere on the forum
let me explain it to you

since a total of 8 tickets is to be distributed to either Jane or Bill, 7 tickets can be distributed either to Jane or to Bill in
(1 + 1)*(1 + 1)*(1 + 1)*(1 + 1)*(1 + 1)*(1 + 1)*(1 + 1)

= 2*2*2*2*2*2*2 = 2^7 ways, and the arrangements will look like as under

0, 7
1, 6
2, 5
3, 4
4, 3
5, 2
6, 1
7, 0

here, as can be seen, in 2^7 ways, one person is getting even number of tickets and the other is receiving odd number of tickets, so

the remaining 1 ticket can be given to the one receiving odd number of tickets in only 1 way to entail the optimum arrangement where each buddy gets even number of tickets

so the answer is (2^7) * 1 = 128

hope this helps and is clear!
thanks

cheers, and do consider some kudos, guys

Rw27

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03 Feb 2018, 02:17

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Shouldn't each case be multiplied by, since they are different tickets?

KarishmaB

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03 Feb 2018, 05:07

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Rw27

Shouldn't each case be multiplied by, since they are different tickets?

Note that each ticket is different hence we have 2 ways of distributing each ticket (except the last one)

Ticket A can be given in 2 ways.
Ticket B can be given in 2 ways.
Ticket C can be given in 2 ways.
and so on...

Note that they don't have to be arranged in a row. If Jane gets 4 tickets A-B-C-D, it is the same as getting B-D-C-A.

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21 Feb 2018, 06:21

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VeritasPrepKarishma

Rw27

Shouldn't each case be multiplied by, since they are different tickets?

hi mam
feeling great to see you here. So far I can remember the solution I have explained here was actually given by you

anyway, I have a question to you
if the question asks us to find out the number of ways in which both friends get odd number of tickets, the answer will remain the same, 128 ways, that's okay, but

what to do if the question asks an arrangement where one buddy gets odd number of tickets and another gets even number of tickets?

its never possible, as the total number of tickets is 8. For this question to be answered, the total number of tickets will have to be an odd number.

is that okay?

thanks in advance, mam

KarishmaB

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gmatcracker2018

VeritasPrepKarishma

Rw27

Shouldn't each case be multiplied by, since they are different tickets?

Yes, 8 tickets cannot be distributed in a way such that one gets an odd number of tickets and another gets an even number of tickets. That would add up to an odd number.

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VeritasPrepKarishma

Rw27

Shouldn't each case be multiplied by, since they are different tickets?

many many thanks to you, mam

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18 Mar 2018, 09:58

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Bunuel

pochcc

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.

Hello,
is the following a correct way to think about the solution?

we have eight tickets to distribute to two persons in even numbers.
Total ways of distributing 8 tickets to two persons is 2^8 (an empty subset is allowed since 0 is even so no need to subtract 1)
even numbers of distributions=odd numbers of distributions thus by symmetry the answer is 2^8/2=256/2=128.

can you guys please confirm this approach?

KarishmaB

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afa13

Bunuel

pochcc

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.

How do you figure "even numbers of distributions = odd numbers of distributions" ?

If anything, even is 0-8, 2-6 and 4-4 while odd is 1-7 and 3-5.

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VeritasPrepKarishma,
Obviously I missed that and got lucky. Thank you for pointing it out. So the proper solution is with combinations of 0-8, 2-6 and 4-4.

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Bunuel

Since total tickets =8 (even)
each will receive either even or odd number of tickets
Hence Number of ways = 2^7/2 = 128
D

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Try to give a clear explanation of my approach (although yet mentioned above).

Ways to assign:

8 - 0 ---> 2 ways (8,0 or 0,8)
6 - 2 ---> 2 ways (6,2 or 2,6) but the 6 can be combined in 8! / (6! * 2!) , hence ----> 2 * 28 = 56 #
4 - 4 ---> 1 way but the 4 can be combined in 8! / (4! * 4!) , hence ----> 70 ##

# you are not interested in the other 2 because they are automatically distributed to the other person from the remaining
## the same

For this last concept there is an useful post here

BillyZ

Bunuel

Mel and Nora share a total of three red marbles, two green marbles, and one blue marble. In how many ways can Mel and Nora divide the marbles between themselves, if it is not necessary for each of them to get at least one marble?

A. $6$
B. $18$
C. $24$
D. $36$
E. $72$

Official solution from Veritas Prep.

Combinatorics questions, like this one, so often come down to a matter of approach. That is, we can tell the story of this problem in different ways, each of which leads to a valid solution, but some of which lead to far easier solutions than others.

For instance, we might view this problem as “Mel could get zero, one, two, three, four, five, or six marbles, in a variety of color combinations” and then calculate/count the number of possible combinations for each possible number of marbles. We might also rely on the symmetry between Mel and Nora to cut that workload almost in half (calculate for Mel getting zero through two, multiply by $2$ to account for Nora getting zero through two, then add the case of Mel and Nora each getting three). But even then, we’d be expending a fair amount of brute force effort.

If we tell the story a bit differently, though, we can make the math a lot better. Rather than focus on one of the people and their number of marbles in the aggregate, focus on one of the colors and the number of that type that one person gets. When it comes to red marbles, Mel could wind up with none, one, two, or three – that’s $4$ possibilities. As for green marbles, Mel could have none, one, or two – $3$ possibilities. And the blue marble is a take-it-or-leave-it proposition – $2$ possibilities. (We don’t have to calculate anything for Nora; she’ll just get whatever is left.) Since we’re distributing red, blue, and green, we multiply these results to find out that there are 4∗3∗2=$24$ total options.

The correct answer is $C$.

Did you notice this problem’s connection to the Unique Factors Trick? What we’re doing here is actually a perfect analogy – we’re taking some quantities of identical items from each of several categories, and it really makes no difference whether those items are various colored marbles or various prime factors. It makes sense that the formula is the same – the number of items in each category plus one (to cover the “choose zero” possibility), and then multiply the results.

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