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There are 4 suits (spades, hearts, diamonds and clubs) and 13 cards for each suit in a deck of 52 cards. A player randomly draw 13 cards. We do not distinguish the cards 1, 2, ..., Q, K.

(a) Find the total number of combinations of suits for the player.

(b) Find the number of combinations of suits that the player has 6 clubs.

(c) Find the number of combinations of suits such that the player has 4 or more spades and 3 or more hearts.

I tried to tackle this question using multinomial distribution but I actually don't really know how to solve it.

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  • $\begingroup$ "We do not distinguish the cards 1, 2, ..., Q, K"? What does that mean? $\endgroup$
    – barak manos
    Commented Feb 15, 2016 at 8:06
  • $\begingroup$ That means that the digits on the cards are ignored, only suits are counted. $\endgroup$
    – Simon Wong
    Commented Feb 15, 2016 at 8:08
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    $\begingroup$ For (a), you need to count the number of ways to write $13$ as a sum of $4$ non-negative numbers. $\endgroup$
    – barak manos
    Commented Feb 15, 2016 at 8:10
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    $\begingroup$ So it should be C(16,3)=560? $\endgroup$
    – Simon Wong
    Commented Feb 15, 2016 at 9:15
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    $\begingroup$ Yep!!! And equivalently for (b), you need to count the number of ways to write $13-6$ as a sum of $4-1$ non-negative numbers. But please wait for other users here to verify this. $\endgroup$
    – barak manos
    Commented Feb 15, 2016 at 9:30

1 Answer 1

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$(a),(b)$ have been resolved in the comments. For $(c)$ you can let the number of clubs, diamonds, hearts, spades picked be $c,d,h+3,s+4$ respectively. So we have $$ c+d+(h+3)+(s+4)=13\implies c+d+h+s=6 $$ Number of non-negative integer solutions of the above equation is $$\binom{6+4-1}{4-1}=\binom{9}{3}=84$$ which is the required number of combinations.

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