Probability that in bridge game the Players N,E,S,W have a,b,c,d spades respectively.

Question

There are 52 cards in bridge and 13 cards of each suit. The formula for numerator is: $${13\choose a}{39 \choose 13-a}{13-a\choose b}{26+a\choose 13-b}{13-a-b\choose c}{13+a+b\choose 13-c}$$

But i don't understand why must we add back a and a+b for the rest cards(i mean $${26+a\choose 13-b}$$ and $${13+a+b\choose 13-c}$$

I thought that we have dealt this spades for other players. Explaint me please this formula Thank you

Answer 1

The first player has taken $13-a$ non-spades from the total of $39$ non-spades to choose from. This means the second player must choose $13-b$ non-spades from a remaining total of $39-(13-a)=26+a$ non-spades.

Likewise, the third player must choose $13-c$ non-spades from the $26+a-(13-b)=13+a+b$ non-spades remaining.