Answer should be 80
4 should be the first element in any sequence without any ambiguity.
Now we have 6 slots where in all slots 2 should come before 1 and 3, and 6 should come before 5 and 7.
Now considering the positions of only 2, 1 and 3 in the insert sequence.
2 would have 4 places out of 6
Say 2 is in the 4th position
Out of the positions indicated by ‘*’, only the last two positions are valid for 1 and 3
thus the total number of such combinations is 2P2. Where P means permutation and mPn means $\frac{m!}{(m – n)!}$
Now considering all the other combinations
Here we would have 3P2 combinations
And so on an so forth.
Thus the total number of sequences = 5P2 + 4P2 + 3P2 + 2P2 = 40
Now after filling up the three positions for 2, 1 and 3 we would have only 3 slots left
There will be a least index among the three slots which would be necessarily filled up by 6 and the remaining 2 slots would be filled by 5 and 7, thus giving 2P2 permutations.
Thus total number of sequences = (5P2 + 4P2 + 3P2 + 2P2) * 2P2 = 40 * 2 = 80