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) |