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Explanation:

x+y+z+t+e = 10
then (x-1) + (y-1) + (z-1) + (t-1) + (e-1) = 5.
Set a=(x-1), b=(y-1), c=(z-1), d=(t-1) ,f=(e-1).
a+b+c+d+f=5
(n+r-1)C(r-1)
n=5 , r=5
9C4 = 126

IMO-C
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The word problem version of the above:

In how many ways can we distribute 10 candies (=identical things) amongst 5 children so that each child receives at least one (since we consider only positive solutions) candy?

Firstly, give one to each child --> 5 candies are left
Secondly, distribute 5 left candies among 5 children; any child may get any number of candies, incl. 0
(5+[5-1])C([5-1])=9!/(5!*4!)=9*8*7*6/4*3*2=126

Why [5-1]? Because we need only 4 dots to get 5s line segments (have a look at Ian's post). And then a problem becomes exactly like a COFFEE/SUCCESS/HHHTT, etc. type of problem.
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