How many solutions are there for xyzwf15 if all x y z w f ar

How many solutions are there for x+y+z+w+f=15 if all x, y, z, w, f are all integers greater than or equal to one, i.e., x,y,z,w,f>=1.

Solution

Consider each of x,y,z,w,f to be a box. You can interpret this as adding 15 balls to five boxes such that each box has one ball. Because each box must have at least one ball (the constraint x,y,z,w,f1), we can add one ball to each bin. We now need to add the remaining 10 balls to the 5 bins.using a stars-and-bars argument, we need 4 divisors to separate the boxes. We have 14 objects (10 balls, 4 dividers), so there are ¹⁴C₄ ways to select (x,y,z,w,f).

so answer will be ¹⁴C₄ .