Question

Number of integral solutions to the equation \(x+y+z=21\), where \(x \geq 1\), \(y \geq 3\), \(z \geq 4\), is equal to ___

Hint:

To find the number of non-negative integral solutions to an equation with constraints, transform the variables to eliminate the constraints and use the formula for combinations:

Answer 1

Correct Answer: 105

To find the number of integral solutions to the equation \(x+y+z=21\) with constraints \(x \geq 1\), \(y \geq 3\), and \(z \geq 4\), we first adjust the variables to account for the minimum values:  Let \(x' = x - 1\), so \(x' \geq 0\). Let \(y' = y - 3\), so \(y' \geq 0\). Let \(z' = z - 4\), so \(z' \geq 0\). The equation becomes: \(x'+1 + y'+3 + z'+4 = 21\) Simplifying, we have: \(x' + y' + z' = 13\) We want the number of non-negative integer solutions to \(x' + y' + z' = 13\). This is a classic stars and bars problem, where the formula for the number of solutions is given by the binomial coefficient: \(\binom{n+k-1}{k-1}\) Here, \(n = 13\) (the total to be partitioned) and \(k = 3\) (the number of variables). Thus, we calculate: \(\binom{13+3-1}{3-1} = \binom{15}{2}\) The number of solutions is computed as: \(\binom{15}{2} = \frac{15 \times 14}{2} = 105\) This solution count fits within the provided range of 105 to 105. Thus, the number of integral solutions is 105.