Counting Problem - Strings What is the number of strings of four decimal digits that contain exactly
one digit repeated twice? (e.g 1198)
My intuition was to first place the digits that aren't repeated and then place the $(10-1)$ remaining digits. Would this be:
$$\binom{10}{2} \times 9$$
Thanks!

Assume that the contraint is exactly one digit is repeated exactly twice. 
STEP 1: Choose the number to be repeated: $10$ options
STEP 2: Choose the other two numbers:
$$\binom{9}{2} \ \ \text{  options.}$$
STEP 3:  Calculate all the permutations of this four characters: $4!=24$
This leads to 
$$10\times 24\times \binom{9}{2} \ \ \text{options,}$$
however, you must divide by $2$ since this distinguish between the repeated digits (e.g. $1189$ is being counted twice as $1\bar189$ and $\bar1189$), thus, the final number is:
$$5\times 24\times \binom{9}{2} $$
