Suppose  x1+x2+x3 +...xn= m where n=3 & xi can have any value of {1,2,3,4,......k } . The task is to find out how many positive integral solution values of xi are there. Here m is also a positive integer. There are 2 situations. If value of k is more than thresh hold limit or equal to thresh hold limit, no. of integral solutions remains unaffected by value of k and this is situation 1. If value of k is less than thresh hold limit , no. of integral solutions depends also on value of k & this is situation 2.

* For situation-1, the result will be correct if n=3 is substituted by any other number but the result for situation - 2 will be erroneous.

SITUATION-1 :  k > = m-(n-1)

Here no. of solutions is C(m-1,n-1)

SITUATION-2 : k  <  m-(n-1)

Here no. of solutions is C(m-1,n-1) -C(k-1,n-1)
n
m
k
Thresh hold Limit of k
This is situation
 
No. of positive integral solutions : C(m-1,n-1) ->                 situation 1
C(k-1,n-1)
No. of positive integral solutions :                                        situation 2