CPhill

Melody has assumed the b***s and the boxes are both indistinguishable....

But.....the question clearly states that they are both distinguishable....capable of being specificallly identified....

Thus. all the b***s are different and all the boxes are different....

(???)

There are 12 parts in the $72

Mahesh gets (7/12)  = $42

Jayrag gets (5/12) = $30

My first answer was c**p....I'll try again

If the b***s and boxes are distinguishable

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We can put all 6 b***s into any box.....there are 3 ways to do this....put them into the first box, the second box or the third box

So 3 ways in this arrangement

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We can put 5 b***s into one box, 1 in another and 0 into the last

 But since the b***s are distinguishable, we can pick any 5 of the 6 to put into a box and we have 3 ways to choose that box. =  3C(6,5) = 18 ways

And we have 1 way to choose the next ball and 2 ways to select the next box = 2 ways

The last box is empty and determined by default

So 18 x 2 = 36 ways in this arrangement

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And we can put 4 b***s into one box and 1 in each of the remaining two.

So we have 3 ways to choose the first box and we want to select any 4 of the 6 b***s to put into that box. So 3C(6,4) = 45 ways

And we have 2 ways to choose the next box and 2 ways to choose the next ball = 4 ways

And the final box and ball are determined by default

So 45 x 2 = 180 ways in this arrangement

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And we can select 3 b***s in one box, 2 into another and 1 in the last

We have 3 ways to select the first box and we want to choose 3 of the 6 b***s to put into that box......so we have 3C(6,3)  = 60 ways to do that

Then, we have 2 ways to choose the next box to put 2 of the remaining 3 b***s into....so this is  2C(3,2)  = 6 ways

Again the last ball and and box are determined by default

So the total ways here are  60 x 6 = 360 ways in this arrangement

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And we can put 4 b***s into one box and two into another

We have 3 ways to choose the first box and we want to choose any 4 of the 6 b***s to put into this = 3C(6,4) = 45 ways

And we want to put the other two b***s into either of the remaining two boxes = 2 ways to do this

The last box is empty by default

So 45 x 2 = 90 ways in this arrangement

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And we can put 3 b***s into one box and 3 into another

There are 3 ways to choose the first box and we want to choose 3 of the 6 b***s

So 3C(6,3) = 60 ways

And there are 2 ways to select the next box to put the remaining 3 b***s into = 2

The last box is empty by default

So 60 x 2 = 120 ways in this arrangement

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And finally, we can put 2 b***s into every box

We have 3 ways to choose the frst box and we want to choose 2 of the 6 b***s to put into that box  3C(6,2)  = 45 ways

And then we have 2 ways to select the next box and we want to choose 2 of the remaining 4 b***s to put into this  = 2C(4,2) = 12 ways

The last 2 b***s and box remain by default.

So 45 x 12 = 540 ways

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So, to recap

6 0 0 =  3 ways

5 1 0 = 36 ways

4 2 0 = 90 ways

4 1 1 = 180 ways

3 3 0 = 120 ways

3 2 1 = 360 ways

2 2 2 = 540 ways

So   3 + 36 + 90 + 180 + 120 + 360 + 540 = 1329 ways

Thanks for that link, Anonymous.....that really helped me see the logic behind the formula!!!

I'm gonna' be honest, Anonymous....I "borrowed" this "formula" from one of our fellow posters, Nauseated, based on his answer to a problem just last week....I haven't had time to digest the rationale behind it, yet.

However.......here is his "non-technical" explanation.....if you care to look at it....

Thanks, Damian......that was a good problem, too.....a little bit of a challenge.....!!!!

We have n ways to choose 1 person from a department with n people.

Then, from the total remaining  people - 83 - , we just want to choose any 2 of them.

So we have....

25 x  23  x 14 x 7 x 19 x C(83, 2)  = 3,643,421,950 possible teams