# Permutation and Combination

• Jul 7th 2009, 08:47 AM
pankaj
Permutation and Combination
There are 5 mangoes,10 apples and 15 oranges.Find number of ways to distribute 15 fruits to two persons.
• Jul 7th 2009, 08:58 AM
Plato
Quote:

Originally Posted by pankaj
There are 5 mangoes,10 apples and 15 oranges.Find number of ways to distribute 15 fruits to two persons.

There are $\binom{X+2-1}{2-1}=(X+1)$ ways to give X identical items to two people.
So what is the solution to your question?
• Jul 7th 2009, 08:59 AM
malaygoel
There are 30 fruits

And does two mangoes are different from each other...or they can be assumed identical?
e.g. first mango is same as second mango?????
• Jul 7th 2009, 09:02 AM
Random Variable
How about expanding $\Big((1+x+x^{2}+...+x^{5})(1+x+x^2+...+ x^{10})(1+x+x^2+...+x^{15})\Big)^{2}$ and finding the coefficient of $x^{15}$ ?
• Jul 7th 2009, 09:53 AM
Random Variable
11304?
• Jul 7th 2009, 06:19 PM
pankaj
Quote:

Originally Posted by Plato
There are $\binom{X+2-1}{2-1}=(X+1)$ ways to give X identical items to two people.
So what is the solution to your question?

But before that you need to select $15$ fruits.And then when you distribute them all fruits are not identical i.e there are $3$ different varieties of fruits.Your formula works only when all objects are alike and of same genre.

Quote:

Originally Posted by Random Variable
How about expanding $\Big((1+x+x^{2}+...+x^{5})(1+x+x^2+...+ x^{10})(1+x+x^2+...+x^{15})\Big)^{2}$ and finding the coefficient of $x^{15}$ ?

This would be number of ways to select $15$ fruits from given $30$ fruits and comes out to be $66$.
Now,what is to be done.

Unless we take into account each selection separately then only can the answer be determined.
• Jul 7th 2009, 06:28 PM
Random Variable
It's not 66. I squared the entire quantity. If there were n people, I would have rasied it to the nth power.
• Jul 7th 2009, 06:32 PM
pankaj
15 fruits from available 30 fruits are to be distributed among 2 persons. It is not that one person gets one set of 15 fruits and the other gets a different set of 15 fruits.

For malaygoel

Fruits of the same type are identical
• Jul 7th 2009, 06:50 PM
Random Variable
nevermind
• Jul 7th 2009, 07:19 PM
billym
Would it be

[30!/(5!10!15!)]*C(30,15)*C(16,15)

Like... the number of permutations of the 30 fruits mulitplied by the number of combinations of size 15 from 30 possible fruits mulitiplied by the number of different ways you can distribute 15 fruits to 2 people...

Way off?
• Jul 7th 2009, 07:50 PM
billym
oops
• Jul 7th 2009, 08:08 PM
malaygoel
Quote:

Originally Posted by pankaj
There are 5 mangoes,10 apples and 15 oranges.Find number of ways to distribute 15 fruits to two persons.

let us pick a collection of 15 fruits:
2 mangoes, 7 apples and 6 oranges.

now, these can be divided among two people in 3*8*7=168 ways
(these include the cases in which one of them don't get anything).

Let us say that this case corresponds to $(3x^2)(8x^7)(7x^6)$

Then all the solution will be given by $\sum [(n+1)x^n][(m+1)x^m][(16-n-m)x^{15-n-m}]$...subject to constraints:

$0 \leq n \leq 5$

$0 \leq m \leq 10$

If the reasoning till now is true, then the solution of given problem is the coefficient of $x^{15}$ in
$(1+2x+...+5x^4+6x^5)(1+2x+...+10x^9+11x^{10})(1+2x +...+15x^{14}+16x^{15})$
• Jul 7th 2009, 08:44 PM
Random Variable
Yeah, my approach would only work if all the fruits were the same. (Doh)