How many permutations of [9] have no adjacent odd digits?

Oct 2009
255
20
St. Louis Area
How many permutations of [9] have no adjacent odd digits?

First look at the same problem for [5] we have:
12345 12543 14325 14523
32145 32541 34125 34521
52341 52143 54321 54123

This suggests that for each odd digit we have a permutation of the rest of the even digits times the permutation of the rest of the odd digits (I think) giving

\(\displaystyle
3*\left(2\right)_{2}*\left(2\right)_{2}
\) = 12

Then the answer to the original question would be

\(\displaystyle
9*\left(4\right)_{4}*\left(4\right)_{4}
\) = 5184

Does this solution hold any merit or am I totally confused as usual?

Thanks
 

Plato

MHF Helper
Aug 2006
22,455
8,631
Notice that in [9] there are five odd numbers and four even.
The even 'separate' the odds: "_2_4_6_8_". Fill in the blanks with the odds.
There (4!) ways to arrange the evens and then (5!) ways to arrange the odds.
SO?
 
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