Question

no of ways x + y + z = 12

Answer 1

91

Answer 2

The problem asks for the number of non-negative integer solutions to the equation $x + y + z = 12$.

This is a classic combinatorial problem that can be solved using the stars and bars method. We have a sum of 12 (the "stars") to be distributed among 3 variables (the "bins"). The number of non-negative integer solutions to the equation $x_1 + x_2 + \dots + x_k = n$ is given by the formula $\binom{n+k-1}{k-1}$.

In this case, $n = 12$ (the sum) and $k = 3$ (the number of variables $x, y, z$). Using the formula, the number of solutions is:

$$\binom{n+k-1}{k-1} = \binom{12+3-1}{3-1} = \binom{14}{2}$$

Now, we calculate the value of $\binom{14}{2}$:

$$\binom{14}{2} = \frac{14!}{2!(14-2)!} = \frac{14!}{2!12!} = \frac{14 \times 13}{2 \times 1} = 7 \times 13 = 91$$

Thus, there are 91 non-negative integer solutions to the equation $x + y + z = 12$.