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 6 points if you must draw exactly 4 “low” cards?	Pick one
A) 1 B) 2 C) 3 D) 4 E) 5
Great question Ravih. This is a permutations problem (order matters) with repeating elements. Given thatlowcards are worth 1 pt andhigh cards2 pts, and you must draw 3 low cards, we know that you must also draw 1 high card. The formula for permutations problems with repeating elements isN!/A!B!...where N represents the number of elements in the group and A, B, etc. represent the number of times that repeating elements are repeated. Here there are 4 elements and thelowcard is repeated 3 times. As a result, the formula is:   5!/4! which represents (5*4*3*2*1)/(4*3*2*1) which simplifies to just 5, giving you answer 5.
<result>E</result>
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 6 points if you must draw exactly 4 “low” cards?