In how many ways 8 different tickets can be distributed

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.

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.

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

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

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?

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.

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.