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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 $=5!\cdot42$.

aarbee
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1 Answers1

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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 $4$. $$-\bullet-\bullet-\bullet-\bullet-\bullet-$$

Christian Blatter
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  • Hey, I didn't understand when you said 'at most one even digit can be placed'. Shouldn't we say at most 3 even digits? – aarbee May 22 '20 at 16:42
  • @Ramit: No two even digits are allowed to be adjacent. – Christian Blatter May 22 '20 at 18:59
  • When you say 'these digits produce 6 spaces in between them', are you talking about odd digits? But odd digits are five. So, they just have 4 spaces in between them. If you are talking about all the digits then they have 7 spaces in between them. If you are talking about the positions of the 8-digit numbers then why are we saying even should be at the ends, they can be in the middle as well? – aarbee May 23 '20 at 04:32
  • Thanks for the edit. I am able to solve the question now. – aarbee May 23 '20 at 08:26