Given that $a+b=4$ and $a^{2}+b^{2}=6$, we want to find the value of $ab$.

Starting with the given equations:

1. $a+b=4$

Squaring both sides of this equation gives us:

$$(a+b)^{2} = 4^{2}$$

Expanding the left side, we get:

$$a^{2} + 2ab + b^{2} = 16$$

2. We are also given that $a^{2} + b^{2} = 6$. Substituting this into the expanded equation, we have:

$$6 + 2ab = 16$$

Solving for $ab$, we subtract $6$ from both sides:

$$2ab = 16 - 6$$

$$2ab = 10$$

Dividing both sides by $2$ gives us:

$$ab = \frac{10}{2}$$

$$ab = 5$$

Therefore, the value of $ab$ is $\boxed{5}$.