There are $13$ "interletter gaps" in SesameStreet, of which $11$ are genuinely between letters, and the other $2$ are at the two ends. We can either (i) choose $1$ of these gaps, and put in it any string of two digits or (ii) choose $2$ of these gaps, and insert a digit into the leftmost chosen gap, and then a digit into the other gap.
For type (i), there are $\binom{13}{1}$ ways to choose the gap, and for each such way we can enter any of the $(10)(10)$ sequences of two not necessarily distinct digits.
For type (ii), there are $\binom{13}{2}$ ways to choose the two gaps, and for each way there are $(10)(10)$ ways to fill the two gaps with digits.
So we get a total of
$$\binom{13}{1}(100)+\binom{13}{2}(100)$$
ways to do the job. Calculate. We have $\binom{13}{1}=13$ and $\binom{13}{2}=78$. This gives a total of $9100$.