Question
Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?

(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040

In the solution I very much like the approach of treating the 5 donuts and 2 dividers as 7 objects leading to the solution of 7! / (2! * 5!) = 21

However, my question is, how could you manipulate that short-cut if the problem stated that each man, had to have at least 1 donut (ie, 1-3 as opposed to 0-5)?
My guess is, since each man has to have at least 1 donut, that leaves only 2 donuts to work with, and 2 dividers, so answer is 4! / (2! * 2!) = 6
Is that right?

Yep, that sounds right. The condition that each man should have at least one donut effectively takes three donuts out of play.

The answers 21 & 6 R correct under the assumpltion that all the donuts are indentical and indistinguishable in all respects, If NOT its another matter!!!

What are the two deviders? (the solution looks like choose 2 of 7).
I did not get this approch can you please explain?

the basic idea is that you have two 'dividers' (|) and five 'donuts' (0).
this is a slick way to think about the problem: anything to the left of the first "|" is larry's, anything between the two "|"s is michael's, and anything to the right of the second "|" is doug's.

so, for instance,
0|00|00 = 1 donut for larry, 2 for michael, 2 for doug
||00000 = doug gets all 5 donuts.

i don't think we'd realistically expect you to come up with a strategy / representation like this on your own, but, now that you've seen it here, keep it in mind in case you ever encounter another 'free distribution' problem like this one.

Free distrobution formula of 'n' objects to 'r' pockets is meaning (5 + 3 - 1 ) = (7).
( 3 -1 ) = (2)
[/code]

For something small like that image, please just type it out. The images take up a lot of space and take time to view. Thanks!

How is this problem different from a scenario of 3 letterboxes and 4 letters wherein the solution is 3 pow(4) .

two things:
1 * please explain the "letterbox" problem in more detail; i don't think i understand what you're saying are the details of the problem.
2 * what is "pow"? is that supposed to mean 3 to the 4th power?