Quote:
Bunuel,
Why is the same formula \(n+r-1C_{r-1}\) used for number of ways of dividing n identical items among r persons whom can receive can receive 0,1,2 or more items . Are you saying the same formula is used even if each person can receive 2 or 3 or anyitem? Why isn't that being factored in to the formula?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 1 mango?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 2 mango?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 3 mango?
Are you saying this formula gives us the same answer for all these questions? (10+4-1)C(4-1)?
I guess you already don't need a reply, but nevertheless:
1) 10 mangoes to 4 kids:
We have 10 mangoes and 3 separators, so together 14 elements, then the number combinations is : \(\frac{13!}{3!10!}\)
2) 10 mangoes to 4 kids, if each one has to get at least 1 mango:
First we give 1 mango to each, this leaves 6 mangoes, so we need to distribute this 6 mangoes to them => 6 mangoes and 3 separators: \(\frac{9!}{6!3!}\)
3) 10 mangoes to 4 kids, if one has to get at least 2 mangoes:
First we give 2 mango to each, this leaves 2 mangoes, so we need to distribute this 2 mangoes to them => 2 mangoes and 3 separators: \(\frac{5!}{2!3!}\)
4) In order for each one to have 3 mangoes, the total number of mangoes has to be 12, so this option is not feasible