Are you smarter than me? 19

$s(n) =k, s(k) = 2, k \in \{ 20, 11, 2\}$

$I) s(n) = 20 = k ,$

If The Number n's digits have No Order, we can select these elements.

$\{992, 983, 974, 965, 884, 875, 866, 776 \}$

so, $3! \times 4 + \sideset{_3}{_2}C\ \times 4 = 36$

$II) s(n) = 11 = k ,$

$\{119, 128, 137, 146, 155, 227, 236, 245, 335, 344 \}$ and $\{920, 380, 470, 560\}$

$3! \times 5 + \sideset{_3}{_2}C\ \times 5 + 4\times2\times2 = 45 + 16$ $=61$

$III) s(n) = 2 = k ,$

$\{20, 11, 2 \}$

$3$

Now, we can get $36 + 61 + 3 = 100$