Mathematics - Permutation & Combination Question with Solution | TestHub

Only one possibility is there, $(\mathrm{9,1},2)$ or $(\mathrm{0,5},7)$ as in even or odd places
$\therefore$ Total number of required $6$ digit numbers

$=(9,1,2$ in odd and $0,5,7$ in even places$)$$+$$(9,1,2$ in even places and $0,5,7$ in odd places$)$
$=3!\times 3!+\left(3! –2\right)\times 3!$
$=36+24$
$=60$