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Instant Math
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Can anyone confirm my answers for a series of counting problems?
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Azaler

by Azaler at March 11, 2010

So I got a bunch of counting problems for the permutations and combinations section, and I'm not really sure if I got them right. If you answer my question, can you tell me your answers and processes (or if they're right at least)?

So, here are the problems.
Find the number of 5-card hands that contain the cards specified.
1. 5 face cards (jacks, queens or kings)
I got 792 with 12 C 5.
2. 4 aces and 1 other cards
I got 48 with 4 C 4 + 47.
3. 2 aces and 3 kings
I got 16 with 4 C 2 + 4 C 3.
4. 4 of one kind and 1 of a different kind ("4 of one kind" means the same value of cards, i.e. 4 kings or 4 7's)
I got 25 with 12 + 13 (no idea if there's a "proper" way to set it up).


Thanks in advance!

Answers

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1. you are correct 2. Your answer is correct but the method is wrong. 4C4 x 48C1 Remember that "and" problems are solved using multiplication and "or" problems are solved using addition. There are 48 cards that are not aces and you need one. 3.  4C2 x 4C3 = 6 x 4 = 24 4.  There are 13 different "kinds" of cards. ones, twos, threes, etc. There are only 4 of each kind. After you pick your "4 of one kind" you have 12 different kinds left to get your one from. 13 x 4C4 x 12 x 4C1 = 624 You could also use this reasoning to find a full house- 3 of one kind and 2 of another. 13 x 4C3 x 12 x 4C2
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Nancy095 Nancy095 March 12, 2010

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