in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?.............njoy
Unless I'm understanding something wrong, in which case tell me:
Op 1: Choose something for first person: 12 ways
OP 2: Choose something for second person: 11 ways
...
Op 10: Choose something for tenth person: 3 ways
Op 11: Choose people for the 2 remaining fruits: 6^2 ways
Total: 12! * 6^2 / 2 = 18 * 12!
Other way:
Op 1: Assign a fruit for each person: 12P10
Op 2: Choose person for remaining fruit: 6^2
I do think that you have misunderstood this problem.
In your solution, you have assumed that the pieces of fruit are all distinct.
If that is the case, then the problem is trivial; just count the number of surjections from a set of 12 to a set of 10: Surj(12,10)=6187104000.
But if we assume that all the apples are virtually identical as well as the oranges are virtually identical, then this becomes an interesting counting problem. Each person gets at least one piece of fruit. It makes a difference is person A gets an apple, an orange, one of each, or two of the same. That is true of each of the other nine people.
Do you want to rethink your solution?
The answer is at least $\displaystyle 10!$ because 10 pieces of fruit will be split up evenly among the 10 people. Now, there are 2 pieces of fruit left, and order doesn't matter for how these pieces are to be split up among the 10 people. This is simply a combination: $\displaystyle 10\choose2$ $\displaystyle =45$, but remember the two pieces of fruit can be two oranges, one apple and one orange, OR two apples. This is four different ways those two pieces of fruit can be split up -- remember the one apple and one orange can switch spots and be a different set.
$\displaystyle 10! + 45(4) = 3628980$
Colby2152, how did you get (10!)? Are you assuming 12 different pieces of fruit?
Did you read my response to the first solution? Assuming 12 different pieces of fruit, makes this a trivial problem.
The only interesting problem is to assume that the apples are virtually identical as are the oranges.
True, what was I thinking. We have six oranges and six apples. Each person can have an apple or an orange, so that is $\displaystyle 2^{10} = 1024$ possibilities. The two pieces of fruit that are left can be split up ten different ways if both pieces goto one individual, but the two pieces can take on 3 different forms giving us 30 possibilities. If the pieces of fruit are split up, then there are $\displaystyle 10\choose2$$\displaystyle *4 = 180$ ways the fruit can be split up among two people. There are four ways that a pair of fruits can look like:
Apple, Apple
Apple, Orange
Orange, Apple
Orange, Orange
I explained that combination in my last post. Now, let's sum the possibilities:
$\displaystyle 1024(30+180)=215040$
EDIT: Somehow, I do not trust this answer considering the first ten pieces of fruit can be evenly split apples and oranges or be a six and four split.
This one of those multiple case problems.
In one case, someone could get three pieces of fruit: three apples, three oranges, two apples & an orange, or two oranges & an apple. Each of those changes the other distributions.
Or two people could get two each with a similar set of sub-cases.
Do you want to see what you can do with that?
Was just wondering, no idea whether this is right or wrong...
Say the total number of solutions is S,
Lets label all the oranges and apples so they are distinct, now the problem reduces to distributing 12 fruits among 10 boxes. So we have $\displaystyle 21 \choose 9$ solutions. If the total number of solutions is S, we can arrange the apples in 6! ways and similarly for oranges, once we remove labels. So
$\displaystyle S \cdot 6! \cdot 6! = {21 \choose 9}$
So $\displaystyle S = \frac{{21\choose 9}}{(6!)^2}$
I hope I am right
Whoops!
My discounting of solutions was wrong
I cannot divide by 6!
Lets try a different approach...
Can I say the event of distributing apples and oranges are separate?
I mean I will count the number of ways to distribute apples only and then, for each such distribution I can have all of "oranges distribution".
I can give 'x1' number of apples to the first guy, x2 to second and so on.
So I am asking total number of non-negative solutions to
x1+x2+x3+.....+x10 = 6.
The total number of solutions for this $\displaystyle {10+6-1\choose 6-1}$. Similarly for oranges
So total solutions is $\displaystyle ({15\choose 5})^2$
Now, how is it??
P.S: How do I get a large bracket around that last square number in LaTex??