0
$\begingroup$

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 $C(26,8)/C(52,26)$ ?

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 2nd problem.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

  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.

$\endgroup$
3
  • $\begingroup$ Thanks! 1) if interest in the probability, then $C(4,2)C(26,8)/C(52,13)$? $\endgroup$
    – itproxti
    Dec 19, 2016 at 19:39
  • $\begingroup$ You should divide by $\binom{52}{8}$. $\endgroup$
    – angryavian
    Dec 19, 2016 at 20:21
  • $\begingroup$ 2) you're missing $C(4,3)$ 3 suits out of 4. $\endgroup$
    – itproxti
    Dec 20, 2016 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.