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I am trying to answer Question 1(d). A valid hand would be Ace hearts, 10 hearts, 2 clubs, 7 clubs, Ace spades, J spades.

Question 1

My attempts have given me 2 different answers that I am unsure of.

What I did was I had to choose 3 suits from 4, then choose 2 cards from each of those suits. When choosing the 2 cards from each suit, do I have to do it 3 times? Would it be 4C3 x (13C2)^3 or 4C3 x 13C2

Thank you.

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    $\begingroup$ Yes you have to do it three times, once for each of the three suits $\endgroup$
    – Henry
    Mar 23, 2020 at 23:42

1 Answer 1

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You are counting ways to select: two from thirteen kinds for each of three from four suits.$$({^{13}\mathrm C_2})^3\cdot{^4\mathrm C_3}$$


Note: ${^{13}\mathrm C_2}\cdot{^4\mathrm C_3}$ would count ways to select the same two from thirteen kinds in three from four suits.

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  • $\begingroup$ When you say selecting the same two from thirteen kinds, do you mean the same 2 numbers for all 3 suits? for example {Ace hearts, 5 hearts, Ace clubs, 5 clubs, Ace spades, 5 spades} $\endgroup$ Mar 23, 2020 at 23:55
  • $\begingroup$ Yes, indeed. When allowed to select different kinds, you will have more ways to do so. $\endgroup$ Mar 24, 2020 at 0:04
  • $\begingroup$ Thanks for the explanation! $\endgroup$ Mar 24, 2020 at 0:07
  • $\begingroup$ I have another quick question. I want to know the number of hands with 2 cards from the same suit. I first choose 2 cards from 1 suit giving me 4C1 x 13C2. Then I am unsure how to choose the rest 4 cards. I assume that they are not allowed to be the same suit as the first 2 cards. Would this be 3C1 x 13C4 or 3C1 x (13C1)^4 ? $\endgroup$ Mar 24, 2020 at 0:17
  • $\begingroup$ @fatimahfatcakes Since there must be a suit with at least two kinds in the hand, we should assume that is asking for a suit with exactly two kinds . Then it becomes as easy as PIE (ie, use the Principle of Inclusion and Exclusion).$$\def\C#1#2{{^{#1}\mathrm C_{#2}}}\C 41\C{13}2\C{39}4-\C 42(\C{13}{2})^2\,\C{26}2+\C43(\C{13}{2})^3$$ $\endgroup$ Mar 24, 2020 at 0:33

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