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A card game called “high-low� divides a deck of 52_kapla

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by conquistador » Sat Nov 21, 2015 9:36 am
A card game called "high-low" divides a deck of 52 playing cards into 2 types, "high" cards and "low" cards. There are an equal number of "high" cards and "low" cards in the deck and "high" cards are worth 2 points, while "low" cards are worth 1 point. If you draw cards one at a time, how many ways can you draw "high" and "low" cards to earn 5 points if you must draw exactly 3 "low" cards?

A. 1
B. 2
C. 3
D. 4
E. 5

can someone explain the language of the question for me. I felt like the answer should be:
26C1 x 26C3
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by [email protected] » Sat Nov 21, 2015 11:50 am
Hi Mechmeera,

Some of the information in this prompt is ultimately not a factor in the solution (the fact that there are 26 'high' cards and 26 'low' cards is not a factor - it just establishes that there are enough cards to score "5 points").

We're told that high cards are worth 2 points and low cards are worth 1 point. Then we're told that we score 5 points when drawing exactly 3 low cards. Since those 3 low cards are worth a total of 3(1) = 3 points, the remaining 2 points MUST come from a high card.

Thus, we have 3 low cards and 1 high card. Drawing cards one at a time, there are only a certain number of ways to score 5 points under these conditions:

HLLL
LHLL
LLHL
LLLH

4 ways.

Final Answer: D

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by MartyMurray » Sat Nov 21, 2015 2:13 pm
The point of this question is much simpler than what the prompt seems to indicate.

All you are really asked to figure out is the order of the 4 cards, 3 low and 1 high, drawn.

The question does not really make this clear, but the answer choices do.

One easy way to figure out the answer is to see that the high card can come in pick 1, pick 2, pick 3 or pick 4, with the 3 low cards filling the other picks.

Given that the only difference between the possible card drawing orders is the placement of the pick of the high card, we are basically choosing 1 out of 4 choices.

So the answer can be calculated via 4C1 = 4.

Choose D.
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by Matt@VeritasPrep » Fri Nov 27, 2015 2:44 am
I think the danger here is assuming that the deck is a standard deck of 52 playing cards. If the high cards are all cards that simply say "H" and the low cards are all cards that simply say "L" (i.e. they are not Jacks, Queens, Kings, etc.), then we have 26 identical cards and 26 other identical cards.

Given identical cards, we can't treat all 26 as being distinct, so we only have the four cases noted in the other solutions.
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