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There is a Preferans problem:

South chose crosses as the trump. How many tricks will get South?

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Part of the solution the author of problem proposes and that I can't understand is:

You can take 7 tricks.
Usually the South want to put ace of diamonds on his first move. It is an error, if the East will answer with ace king, then just six tricks for South, no options, the game is simple. On move from the clubs marriage, answer with crosses or diamonds - no difference. After drawing these suits there is move with a spades left, that leads to desired 6 tricks for South. ( If you understand russian you can find solution here: )

Please help me to understand this. What I see as best play for West-East are:

1. (S) Ad - (W) Kd - (E) 9d
2. (S) Qd - (W) 8d - (E) 10d
3. (S) 7d - (W) 9s - (E) Jd

The task for east here to avoid move in hearts in order to win queen of hearts. But east have only 2 cards to give up.

4. (E) 10c - (S) Qc - (W) 10s
5. (S) 7c - (W) Js - (E) Jc
6. (E) Qs - (S) 8c - (W) Ks
7. (S) 9c - (W) 7h - (E) Ac
And east have to give up hearts. That leads to only 3 tricks for them.
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