East/West can make 5 ♠
, losing just the major aces (pitching
a club on the diamond ace.)
North/South can only make six tricks in clubs - four clubs and two aces -
but that's not the limit of the hand. Would you believe North/South
can make nine tricks in diamonds, with the 7-0 split?
Say West leads a major. Which is irrelevant, so say spades. Declarer wins,
and leads high clubs, West ducks one round and wins the second round.
West leads a heart, again won in the South hand, and two clubs are run,
pitching hearts from North, leading to:
||K Q 8 6 4
East's hand is irrelevant, with West holding all trumps.
A heart is led from South, and West must ruff, but how high? It's a losing
proposition to ruff with the ace. A ruff with the ten is overruffed, and
North exits a spade, West forced to ruff. West is on lead at:
West can cash the ♦ A
and exit a diamond, but dummy finesses
the diamond, and exits again in spades, and West is forced again to lead
from his diamond suit.
West could lead diamonds earlier, but there is no way for him to
prevent declarer from scoring four diamonds along with three clubs and
the two major aces.
That's nine tricks, if you're still with me, making the 6 ♦
sacrifice a winner. Since East/West make 11 tricks in their seven-card
spade suit, and North/South make nine tricks in their six-card diamond
suit, that's 20 total tricks on a hand with 13 total trumps.