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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 Feb 15 '16 at 8:06
  • $\begingroup$ That means that the digits on the cards are ignored, only suits are counted. $\endgroup$ – Simon Wong Feb 15 '16 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 Feb 15 '16 at 8:10
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    $\begingroup$ So it should be C(16,3)=560? $\endgroup$ – Simon Wong Feb 15 '16 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 Feb 15 '16 at 9:30
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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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