How many eight-card hands can be chosen from exactly 2 suits/13-card bridge hands contain six cards one suit and four and three cards of another suits

Question

1. How many eight-card hands can be chosen from exactly $2$ suits of an ordinary $52$-card deck? (there are $4$ suits clubs, diamonds, hearts and spades

I think since there are $26$ cards in $2$ suits and eight cards from those $26$ (order does not matter), thus $\displaystyle\frac{\binom{26}{8}}{\binom{52}{26}}$ ?

2. How many $13$-card bridge hands can be chosen from an ordinary $52$-card deck that contain six cards of one suit and four and three cards of another two suits? (there are $4$ suits clubs, diamonds, hearts and spades

I do not understand $2$nd problem.

Answer 1

1. I don't know why you are dividing by $\binom{52}{26}$. The number $\binom{26}{8}$ already counts the number of 8-card hands you could get from a pre-specified $26$ cards. You may need to multiply by $\binom{4}{2}$ to choose which two suits to use, though.

2. There are $4 \cdot 3 \cdot 2$ ways to choose the 6-card suit, the 4-card suit, and the 3-card suit to make up your hand. Then you multiply by $\binom{13}{6}$, $\binom{13}{4}$, and $\binom{13}{3}$ to choose the number of cards of each suit.