In how many ways 8 different tickets can be distributed

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.

Hope it's clear.

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.

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

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.

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.

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

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.

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.

Sorry but I do not understand the answer choices. From x to y ?? is it the number of ways?

Yes. In which range is the correct answer.

Nice.... loved this one!

I solved it but it took too much time!
Surely I would have jumped

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.

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?

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.