Number of ways to distribute $20$ identical pencils to $6$ children with no restrictions. First time posting a question so assistance is greatly appreciated.  I saw a similar question but didn't see the response to item d.) which is what I need to finish a comparable question.  
Number of ways to distribute 20 identical pencils to 6 children with no restrictions.  
d) If the pencils are given out randomly. What is the probability that there are at least two kids receive the same number of pencils if every kid receives at least one pencil?
Please provide a step-by-step answer on how you got your final answer. I used combinations to arrive to the other answers so that will be helpful.  Please let me know if I'm unclear.  Thank you all for your assistance.  If this was already answered, please provide the link of the post.  Thanks!
EDITED:
In a similar question, someone wrote...
If they all get the different number of pencils, then there would be at least 1+2+3+4+5+6=21 pencils which is impossibile.
So the probability that at least two will get the same number of pencil is 1.
It is impossible that everyone gets a different number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%.
I understand that yes, two kids will at least receive the same number of pencils but I don't understand the equation if any.  I drew it out but can't think of the equation.  Would it be C(20,8)?
 A: I think this maybe what you are looking for...
a) Say each of the 6 children were to get one pencil each, then that can be done in only 1 way, since the pencils are all identical, it doesn't matter who gets which one, they are all the same.
or
b) Each of the six children get at-least one pencil.Then this is a stars and bars problem. It's like solving  x1 + x2 + x3 + x4 + x5 + x6 = 20, but all xi take positive integer values > 0.Then the answer will be 20-1C6-1 ways.
or
c) It's possible that at-least one kid gets no pencil, which is the same as solving x1 + x2 + x3 + x4 + x5 + x6 = 20, but xi can also be 0 unlike in the previous case. Then this is 20+6-1C6-1 ways.
d) Now to answer, the probability question, like you said this is 1, because if each student gets a different number of pencils, starting with 1 pencil for the first student, 2 pencils for the second, 3 for the the third and so on...then, the least possible number of pencils such that each student gets a different no: of pencils, will be 21, since, 1+2+3+4+5+6 = 21. Therefore, at-least two students have to get the same number of pencils to make the total 20.
I guess we can write that as p(atleast two students getting same no of pencils) = 1 - p(everyone getting different no: of pencils) and 
everyone getting different number of pencils will be => x1 + x2 + x3 + x4 + x5 + x6 = 20, such that, x1 ≠ x2 ≠ x3 ≠ x4 ≠ x5 ≠ x6, and xi >0.
