Question 9

$Consider\quad each\quad decimal\quad place\quad but\quad the\quad thousands\quad place\quad in{ \quad N }_{ 2 }.\\ If\quad it\quad is\quad filled\quad up\quad by\quad a\quad digit\quad i,\quad then\quad there\quad are\quad 10-i\\ ways\quad of\quad filling\quad up\quad the\quad same\quad place\quad in\quad { N }_{ 1 }.\quad This\quad makes\quad up\\ 1+2+3+...+10=55\quad ways\quad for\quad each\quad digit \quad of { \quad N }_{ 2 } .\\ As\quad the\quad thousands\quad place\quad cannot\quad take\quad the\quad digit\quad 0,\quad there\quad can\\ only \quad be\quad 55-10=45\quad ways\quad of\quad filling\quad it\quad up\quad in\quad { N }_{ 1 }.\quad So\quad the\quad \\ number\quad of\quad ways\quad of\quad forming\quad these\quad two\quad numbers:\\ 45\times 55\times 55\times 55=\boxed { 7486875. } \quad$

Written properly

$\displaystyle \sum_{k=1}^9 k\left(\displaystyle \sum_{n=1}^{10}n\left(\displaystyle \sum_{m=1}^{10}m\left(\displaystyle \sum_{j=1}^{10}j \right)\right)\right)=7486875$