How many triplets (x, y, z) satisfy the equation $x+y+z=6$, where x, y and z are natural numbers?

The correct option is 10.

We are asked to find the number of triplets of natural numbers ($x$, $y$, $z$) that satisfy the equation:
$x+y+z=6$

In mathematics, natural numbers typically refer to positive integers, i.e., $x,y,z>0$ (or $x,y,z\ge 1$).

We can solve this problem using the stars and bars combinatorics method. The number of positive integer solutions to the equation $x_1+x_2+\cdots +x_k=n$ is given by the binomial coefficient:
$\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$

For our given equation, we have $n=6$ and $k=3$ (since there are three variables: $x$, $y$, and $z$).
Substituting these values into the formula:
$\left(\genfrac{}{}{0ex}{}{6-1}{3-1}\right)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)$

Now, we calculate the value of $\left(\genfrac{}{}{0ex}{}{5}{2}\right)$:
$\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5\times 4}{2\times 1}=10$

Alternatively, we can list all the possible triplets ($x$, $y$, $z$) where $x,y,z\ge 1$ to verify the result:
1. If $x=1$:
    • (1, 1, 4)
    • (1, 2, 3)
    • (1, 3, 2)
    • (1, 4, 1)
2. If $x=2$:
    • (2, 1, 3)
    • (2, 2, 2)
    • (2, 3, 1)
3. If $x=3$:
    • (3, 1, 2)
    • (3, 2, 1)
4. If $x=4$:
    • (4, 1, 1)

Counting all the listed combinations, we find exactly 4 + 3 + 2 + 1 = 10 valid triplets.