There is only one runway which will be used by seven flights one by one. P and Q will use the runway neither first nor last. S will use it second after R. M will use it second before O. N will use it after O and before R. Q will use it just before S. Who will use the runway in the middle?

Let there be 7 positions for the flights: 1, 2, 3, 4, 5, 6, 7.
S will use it second after R: This means there is one slot between R and S, i.e., R→slot→S.

Q will use it just before S: This perfectly fills the slot between R and S! So the sequence is R→Q→S.

M will use it second before O: This means M→slot→O.

N will use it after O and before R: This means the order is O→N→R.

Let's chain all these together:

From 3 and 4, we have M→slot→O→N→R.

Since there is exactly one slot between M and O, and N comes after O, let's see if another element can fit. But wait, we already know R→Q→S follows R.

Let's combine the chains: M→slot→O→N→R→Q→S.

Who is left? The only remaining flight is P. P must occupy the empty slot between M and O!

This gives the complete order: M, P, O, N, R, Q, S.

Let us verify the conditions:

P and Q are neither first (M is first) nor last (S is last). This satisfies the condition.

The 7 flights in order are: 1: M, 2: P, 3: O, 4: N, 5: R, 6: Q, 7: S.

The flight in the middle (4th position) is N.