A child has plastic toys bearing the digits 4, 4 and 5.

Given: We have toys with bearing 4, 4 and 5

To Find: Number of 3-digit numbers he can make

The formula used: The number of permutations of $n$ objects, where $p_{1}$ objects are of one kind, $p_{2}$ are of the second kind, ..., $p_{k}$ is of a $k^{\text {th }}$ kind and the rest, if any, are of a

different kind is $=\frac{\mathrm{n} !}{\mathrm{p}_{1} ! \mathrm{p}_{2} ! \ldots \ldots \ldots \ldots \mathrm{p}_{\mathrm{k}} !}$

The child has to form a 3-digit number.

Here the child has two 4's.

We have to use the above formula

Where,

$\mathrm{n}=3$

$\mathrm{p}_{1}=2$

$\Rightarrow \frac{3 !}{2 !}=3$ ways

The numbers are 544,454 and $445 .$

He can make 3 3-digit numbers.