Three integers a , b , and c are chosen such that their sum is 28 . It is given that a 1 , b 2 , and c is a single-digit positive integer strictly greater than 2 . The number of such ordered triplets (a, b, c) that can be formed is

Step-by-step solution

We are given the equation: a + b + c = 28 with constraints a 1 , b 2 , and c is a single-digit positive integer strictly greater than 2 . This means 3 c 9 . First, we find the total number of solutions ignoring the upper bound on c (i.e., assuming c 3 ). Let a' = a - 1 , b' = b - 2 , and c' = c - 3 . Then a', b', c' 0 . Substituting these into the equation: (a' + 1) + (b' + 2) + (c' + 3) = 28 a' + b' + c' = 22 The number of non-negative integral solutions is ²²⁺³⁻¹C_ 3-1 = ²⁴C₂ = 24 23 2 = 276 . Next, we subtract t