Re: [algogeeks] 23 candies among 7 kids

2010-06-26 Thread harit agarwal
ok.. i got it..
it is 29! / 23! 6!

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Re: [algogeeks] 23 candies among 7 kids

2010-06-25 Thread harit agarwal
look at this approach:

a1+a2+a3+a4+a5+a6+a7=23
now it means there are 23 choclates and 6 partitions(+ symbol) and arrange
these 23+6=29 items so it is 29!
now 23 choclates are identical so divide by 23!

so number of ways=29! / 23 !

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Re: [algogeeks] 23 candies among 7 kids

2010-06-23 Thread harit agarwal
it is 29! / 23!

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[algogeeks] 23 candies among 7 kids

2010-06-22 Thread Aarthi Thangamani
Given 23 chocolates and there are 7 kids. How many ways are there to spilt
the chocolates among the 7 kids? All of the chocolates must be distributed
and it can happen that some kids do not get any.

For example: first kid gets 23 chocolates and rest get none is one
possibility. First 6 kids get one each and the 7th kid gets 17 chocolates is
another possibility. How many such ways are there to distribute?

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Re: [algogeeks] 23 candies among 7 kids

2010-06-22 Thread rajat ahuja
x1+x2+x3+x4+x5+x6+x7=23,,,
noe no of solution is,,, of equation is,,
n+r-1C(COMBINATION )r-1 (n identical items to be distributed into R grups)
so ,, ans 23+7-1c7-1;(identical)
if unidentical case,
then each each cholate wud hav 7 options,,
so ,, ans -- 7^23 (different)

On Tue, Jun 22, 2010 at 8:22 PM, Aarthi Thangamani 
aarthi.thangam...@gmail.com wrote:

 Given 23 chocolates and there are 7 kids. How many ways are there to spilt
 the chocolates among the 7 kids? All of the chocolates must be distributed
 and it can happen that some kids do not get any.

 For example: first kid gets 23 chocolates and rest get none is one
 possibility. First 6 kids get one each and the 7th kid gets 17 chocolates is
 another possibility. How many such ways are there to distribute?



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Re: [algogeeks] 23 candies among 7 kids

2010-06-22 Thread ANUJ KUMAR
(2^7-1)^23

On Tue, Jun 22, 2010 at 8:22 PM, Aarthi Thangamani
aarthi.thangam...@gmail.com wrote:
 Given 23 chocolates and there are 7 kids. How many ways are there to spilt
 the chocolates among the 7 kids? All of the chocolates must be distributed
 and it can happen that some kids do not get any.

 For example: first kid gets 23 chocolates and rest get none is one
 possibility. First 6 kids get one each and the 7th kid gets 17 chocolates is
 another possibility. How many such ways are there to distribute?



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Re: [algogeeks] 23 candies among 7 kids

2010-06-22 Thread bala chander
Hi,

Given that u need to split 23 into 6 parts. Its all about placing + with
24 spots( wit 23 1's )(6 ,5...0). So, the answer is

Description :

1 1 1 1 1 1 1 1 1 1 1 1  .. 1
  + + +

1. Sharing with just one means there is no + - 24 C 0
2. Sharing with just two means there is 1 plus - 24 C 1

Sigma (i=6 to 0) 24 C i

PS : Pls check if 0 cases are included in higher numbers as well

Thanks,
Bala

On Tue, Jun 22, 2010 at 7:52 AM, Aarthi Thangamani 
aarthi.thangam...@gmail.com wrote:

 Given 23 chocolates and there are 7 kids. How many ways are there to spilt
 the chocolates among the 7 kids? All of the chocolates must be distributed
 and it can happen that some kids do not get any.

 For example: first kid gets 23 chocolates and rest get none is one
 possibility. First 6 kids get one each and the 7th kid gets 17 chocolates is
 another possibility. How many such ways are there to distribute?



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Re: [algogeeks] 23 candies among 7 kids

2010-06-22 Thread Jitendra Kushwaha
coefficient of x^23 in   (1 + x + x^2 + x^3 + ... + x^23)  ^ 7

i.e.  29C23

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Jitendra Kushwaha
MNNIT, Allahabad

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