Number of integral solutions to the equation x+y+z=21, where x ≥ 1,y≥3,z≥4, is equal to ____________.

Solution

Maths sir · 3 steps

Pay each variable its minimum first, then share out what is left freely among the three.

1. Remove the minimums. Put x = 1 + a, y = 3 + b, z = 4 + c with a, b, c ≥ 0. The equation becomes a + b + c = 21 - 8 = 13.
2. Stars and bars. The number of non-negative integer solutions of a₁ + a₂ + … + aᵣ = n is C(n + r - 1, r - 1). Here n = 13 and r = 3.

   C(13 + 3 - 1, 3 - 1) = C(15, 2) = (15 × 14)/2 = 105

3. So. Each solution (a, b, c) gives exactly one (x, y, z), so there are 105 integral solutions.

The common slip: 66. Using the positive-integer formula C(n-1, r-1) on a + b + c = 13 gives C(12, 2) = 66, but a, b, c are allowed to be zero.

The idea to keep. Lower bounds on variables? Shift each variable by its minimum so every new variable starts at 0, then use C(n+r-1, r-1) on what remains.

Answer from the NTA final answer key. Solution by PrepWiser.