[sapta]

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.

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

[ksr985]

the answer seems correct.

[sapta]

But as I have said earlier, the book had another answer. The process there also seemed correct.(the solution is too long,but if you want to see it, tell me.I will post it)

[xiankai]

it should be 12! x 11! x 10!, to represent the possibilities that the 3 required fruits can be picked from. then multiply it by 9C3 to represent the remaining fruits.

[sapta]

I don't get it ???What's the 12!x11!x10! for? Is it 12C1 x 11C1 x10C1?And why 9C3?

[billnotgatez]

The exclaimation point ! in the formula represents the term "Factorial"

From the dictionary

The product of all the positive integers from 1 to a given number: 4 factorial, usually written 4!, is equal to 24 (1 × 2 × 3 × 4 = 24).

[xiankai]

sorry my mistake, yeah it should be 12C1, 11C1 and so on ><. actually i would have preffered to use 12x11x10 which is more sensible

9C3 is the number of ways 3 objects can be selected, from 9 objects, irrespective of order.

after some careful thought, i think it should be replaced by 9P3. all the fruits are different, my bad ><

so the answer should be 12x11x10x9P3

its time i brushed up my maths :/

[sapta]

But,9P3 would mean that each of the 3 boys would get same no. of fruits,1 of the 9 fruits precisely.(maybe )However the question permits a boy to get a max. no. of 10 fruits. so,distributions like 10-1-1,    9-2-1 etc  don't come in consideration.  


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To be clearer,9P3 would mean the no. of ways you can give 1 fruit (out of of 9 diiff. fruits) to each of the 3 boys.  

[xiankai]

hmm u're right.

my apologies if i've confused u one way or another.

lets replace the 9P3 with just 9+8+7+6+5+4+3+2+1 instead.(i dont know an easier way to express that o_O)

1st boy can receive 1-9 fruits. if the boy receives 9 fruits, only one outcome is possible. should he get 8, 2 outcomes are possible : 8-1-0 or 8-0-1. should he get 7, 3 outcomes are possible, 7-2-0, 7-1-1 or 7-0-2. as to 6,

so all the outcomes would be 9! provided instead of multiply, u use plus

[sapta]

First of all,if a boy gets the remaining 9 fruits,then 3 outcomes are possible(any of the 3 boys can get it).And it will get more difficult to calculate the other possible distributions.

Secondly,how can you be so sure that the no. of distributions will follow a fixed rule like 3 for 10-1-1, so x for 9-2-1 and so on.

Thirdly,I don't get your last line.(can't use quotes )

LASTLY,, I know the correct procedure to solve the problem as I have said earlier. What I want is to know the error in my procedure(1st post)


 

[xiankai]

ok the way i see it, there is no error in your calculations... but either the question is incorrect or the answer book is incorrect.

[sapta]

Well then find the error in the book's solution.Here it is-

"No. of distributions possible(without any restrictions) is
3^12.----------(1)

No. of cases where a boy can get all the 12 fruits=3C1(any of the 3 boys can get them).Alternately,we can say that in 3 cases 2 boys won't get any fruit.-------(2)

now,if a boy gets no fruit then the other two will get them.No. of distributions possible for 2 boys=2^12.Here also a boy (of the 2) can get no fruit in 2 ways.So, no. of distributions where the 2 boys would get 12 fruits with each having at least one fruit = 2^12-2.
          so,1 out of the 3 boys would get no fruit in 3C1(2^12 -2) ways--------------------(3)

From (1),(2) & (3) we get,Required no. of distributions
=Total distrbutions-(2 boys get no fruit + 1 boy gets    
                                                       no fruit)
=3^12-{3+3C1(2^12-2)}
=519156.
           Ans.

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I am starting to think there may be 2 correct answers.  

[xiankai]

thats basic probability for you, even though it may not be accurate. i still prefer your solution

[sapta]

i still prefer your solution  

thanx for that.But do you mean there ARE two correct answers or are you tired of "finding the error"?

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Hey Mitch ,i think i should expect some suggestion from you.You are the moderator of this forum.

[xiankai]

actually i dont like the book's approach. it doesnt really sound right theoretically, and may not work out in real life.

mitch is more adept in chemistry, which is what this site is about