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### Topic: Find the error  (Read 18343 times)

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#### Borek

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##### Re:Find the error
« Reply #15 on: June 05, 2005, 07:11:58 AM »
Hey Mitch ,i think i should expect some suggestion from you.You are the moderator of this forum.

Moderator is like cop - he is here not to fix your car, but to fine you for speeding
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#### GCT

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##### Re:Find the error
« Reply #16 on: June 05, 2005, 04:23:53 PM »
The following question came at a national level exam-

"In how many ways 12 different fruits can be distributed among 3 boys so that each receives at least 1 fruit?"

I solved it as follows-
No. of selection of 3 fruits out of 12=12C3.3 fruits can be distributed among 3 boys in 3! ways.From the remaining 9 fruits each can be distributed in 3 ways. Therefore all the 9 fruits can be distributed in 3^9 ways.
Hence, the answer is 12C3 x 3! x 3^9.

HOWEVER, I LATER FOUND THE PROBLEM WORKED OUT IN A BOOK.THE PROCESS WAS CORRECT BUT THE ANSWER WAS DIFFERENT.

That means my process must be wrong.so, where’s the error? Plz help.

--------------------------------------------------------------

This is a math’s question. Can I post it here? This is a chemistry site after all.

here's one way I think will work

set the first person's rationing of fruit constant...for instance start with one fruit for individual #1

this leaves 11 fruit left to be distributed towards the other two, the maximum each can recieve for any combination is 10, we will need to consider all possibilities...this can soley be accomplished when each individual receiving a possible 1-10 fruit....with individual #1 receiving 1 fruit, the possibilities for #2 is 1-10, we won't need to repeat this for individual #3 since the outcomes are fixed for #3 (for instance, when individual #2 recieve 7 fruits, #3 will recieve 4, identical to whether we started out with #3 recieving 4 and #2 recieves 7).

Thus

#1,#2,#3

1, 1-10, fixed---10 possibilities

2, 1-9, fixed---9

3, 1-8, fixed---8

4, 1-7, fixed---7

5,1-6, fixed---6

6,---5

7,---4

8,---3

9,---2

10,---1

The answer is the sum of 1-10

n(n+1)/2=55 possibilities

#### GCT

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##### Re:Find the error
« Reply #17 on: June 05, 2005, 05:46:02 PM »
Actually, I forgot the fact that there are 12 different fruits.

#### sapta

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##### Re:Find the error
« Reply #18 on: June 06, 2005, 01:50:53 AM »
Quote
THE PROCESS WAS CORRECT BUT THE ANSWER WAS DIFFERENT
.

I see that this may have caused confusion.I mean to say here that both the process & the answer were different.

#### sapta

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##### Re:Find the error
« Reply #19 on: June 10, 2005, 11:50:28 PM »
I found another obvious reason for my answer to be wrong.(is this correct English? )

Total no. of distributions possible(no restrictions)=3^12
But,that's>>3^12.

How can this be possible?

#### Tedjn

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##### Re:Find the error
« Reply #20 on: July 17, 2005, 03:51:46 PM »
I am not the best at math (very bad at probability) and explaning things, but I'll give this a shot.

The book method seems to be correct. They overcounted and then subtracted away the impossible situations dictated by the requirements.

For your method, there is one slight problem with it. Let us consider the first boy, call him Bob. Assume that Bob receives only two fruits (unlucky that he is), and that he has received F1 and F2.

Your solution takes into account two distributions. The first is only to ensure that each boy receives at least one fruit, and the second to distribute the rest of the fruit. Let's call the set of fruit Bob receives from the first distribution S and the set of fruit from the second distribution T.

By multiplying the two distributions together, you have counted

S = {F1}, T = {F2} and
S = {F2}, T = {F1}

as two different situations when they should only be regarded as one.

This may not seem much, but if Bob received three fruits, there would be an overcounting of 3 for each possible way Bob can receive three fruits, i.e. 12C3. For four fruits, there is an overcounting of 3 for each possible way Bob can receive four fruits.

In addition, each time there was a miscount for Bob, there is also a huge miscount for Fred and George also. Furthermore, after all that, we must also count the miscounts for Fred and George.

Therefore, the miscount is quite large. I believe that is the problem in your solution.