If the total number of ways in which $8$-digit numbers can be formed by using all the digits $0,1,2,3,4,5,7,9$ such that no two even digits appear together is $(5!)k$, then $k$ is equal to?
There are three even digits ($0,2,4$) and five odd digits ($1,3,5,7,9$).
If odd digits are taking odd places then number of digits are $5!\cdot4!$.
If odd digits are taking even places then number of digits are $5!\cdot3\cdot3\cdot2\cdot1$.
So, the total number of digits are $(5!)42$.
But the answer is given as $100$.
Edit: Elaborating my reasoning for the above calculations:
For considering odd or even, I am starting from the left of the numbers. i.e. in the number $abcd$, $a$ and $c$ are at odd places, $b$ and $d$ are at even places.
If five odd digits ($1,3,5,7,9$) are taking four even places they can be arranged in ${5\choose4}\cdot4!$ ways. So, the remaining one odd digit and the three even digits can be arranged in $4!$ ways at even places. So, here, total numbers are $5!\cdot4!$.
If the odd digits are being placed at even places, they can be arranged in ${5\choose4}\cdot4!$ ways. So, the remaining one odd digit and the three even digits would occupy odd places but $0$ can't be placed at the first place. So, that place can be filled only in $3$ ways. So, number of ways for filling the third place are $3$, for fifth place $2$ and for the seventh place $1$. So, here the total number of digits are $5!\cdot3\cdot3\cdot2\cdot1$.
So, total ways $=5!\cdot42$.
There are $5$ odd digits, but only $3$ even digits. Therefore they can't just alternate between odd and even.
You can position the odd digits in so many ways. These odd digits $\bullet$ produce $6$ spaces $-$ in between them and at the ends. Each of these spaces can receive at most one even digit. But you are forbidden to place the $0$ in front. Therefore place first legally the $0$, then the $2$ then the $4$. $$-\bullet-\bullet-\bullet-\bullet-\bullet-$$