My friend David Berkowitz played this deal at Jourdans Bridge Club in Boca Raton, Florida:

Vul: N-S Dlr: West	4 3 K Q 9 4 10 6 4 2 A 10 2
8 7 A 3 2 Q J 9 8 7 5 6 5					K J 10 9 6 5 8 5 3 Q 9 8 7
				A Q 2 J 10 7 6 A K K J 4 3

West opened 2, East responded 2 (not forcing) and David as South ended up in 3NT.

West led a spade, won by declarer.

He knocked out the heart ace and that meant he had the following sure winners:

2 spades, 3 hearts, 2 diamonds, 2 clubs. That's 9 tricks, so it was just a matter of overtricks.

What do you think of the club suit?

Assuming declarer plays East for the queen, he can take 3 tricks (lead the ace, then finesse). Can he get 4 club tricks and make many overtricks?

It looks impossible. When declarer wins the A and leads the 10, East will cover, of course. If declarer doesn't lead dummy's 10, East still gets a club trick.

But, look what happened. When West won his A, he continued spades. Declarer won and cashed the red suits to leave:

Vul: N-S Dlr: West -- -- 10 6 A 10 2
-- -- Q J 9 6 5					K -- -- Q 9 8 7
				2 -- -- K J 4 3

What a curious ending. East had to keep all four clubs (else the suit would run). So, David threw East in with the spade. East had to lead a club and voila! -- 4 club tricks for declarer.

Plus 660 was a top board.

Note -- West could have broken up this ending by playing a club when he was in.