Solution :
There are 4 odd digits (1, 1, 3, 3) and 4 odd places(first, third, fifth and seventh).
At these places the odd digits can be arranged in \(4!\over 2!2!\) = 6
Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in \(3!\over 2!\) = 3 ways
Therefore, The required number of numbers = 6 \(\times\) 3 = 18
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