Question

# The number of solutions (x, y, z) to the equation x - y - z = 25, where x, y, and z are positive integers such that x≤40,y≤12, and z≤12 is 87 105 99 101

Solution

## The correct option is C 99 x−y−z=25 where x, y, z are positive integers such that x≤40,y≤12,z≤12,  y+z=x−25……1 Now,  for LHS to have feasible solutions under the given conditions, RHS = (x - 25) can take values from a minimum 2 to the maximum of 15. i.e. 2≤(y+z)≤15 Now, let's consider y+z≤15 y+z+k=15…2, where k is any variable. Putting y=(y′+1),z=(z′+1) in equation 2), we get y′+1+z′+1+k=15y′+z′+k=13 Number of solutions for the above equation =(13+3−1)C(3−1)=15C2=15×142=105 But the above number of solutions contains solutions with y′≥12,z′≥12. Those have to removed. Number of invalid solutions = 2 + 4 = 6 ( 2 solutions for RHS = 13 and 4 solutions for RHS = 12) Hence, final number of possible solutions = 105 - 6 = 99.

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