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ravih
Joined: 15 May 2014
Last visit: 17 Oct 2014
Posts: 5
Given Kudos: 6
Schools: Insead Sept '16 Anderson '17
D
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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
A. 1
B. 2
C. 3
D. 4
E. 5
09 Jun 2014, 21:31
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Great question Ravih. This is a permutations problem (order matters) with repeating elements. Given that "low" cards are worth 1 pt and "high cards" 2 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 is "N!/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 the "low" card is repeated 3 times. As a result, the formula is:
4!/3! which represents (4*3*2*1)/(3*2*1) which simplifies to just 4, giving you answer D.
Another way to think about it that you must have 3 low cards and one high card, and the question is asking you about the different possible orders. The 4 orders are:
H-L-L-L, L-H-L-L, L-L-H-L, and finally L-L-L-H, giving you 4 options.
I hope this helps!!!
4!/3! which represents (4*3*2*1)/(3*2*1) which simplifies to just 4, giving you answer D.
Another way to think about it that you must have 3 low cards and one high card, and the question is asking you about the different possible orders. The 4 orders are:
H-L-L-L, L-H-L-L, L-L-H-L, and finally L-L-L-H, giving you 4 options.
I hope this helps!!!
ravih
Joined: 15 May 2014
Last visit: 17 Oct 2014
Posts: 5
Given Kudos: 6
Schools: Insead Sept '16 Anderson '17
10 Jun 2014, 00:04
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Thank you Brandon for the explanation.
However, I still have some confusion. I believe that this is a combination problem not a permutation one as the question basically asks us to choose cards (1 from the 26 High cards & 3 from the 26 Low cards). The arrangement of the cards is not moot. Therefore the answer should be:
26C1 x 26C3
Following is the solution from Kaplan which I somehow did not find very convincing. Although they have confirmed that this is a combination problem, they have applied the formula to only the 4 cards and not the selection from the entire batch of 52 cards:
"First, we need to figure out how many cards of each type we will need in order to earn 5 points. The "low" cards are worth 1 point. Since the question says we must draw 3 of these "low" cards, we get 3 points from "low" cards. This leaves us with 5 – 3 = 2 points left that we need to earn with the "high" cards. Since "high" cards are worth 2 points each, if we get one "high" card, we will reach our goal of 5 points total. Hence it takes us four cards to get 5 points: 3 "low" and 1 "high."
Our task is now simply to discover how many ways we can draw these 4 cards. Let’s draw out the possible ways: (L L L H), (L L H L), (L H L L), (H L L L) for a total of 4 different ways to get 3 "low" and 1 "high." Answer Choice (D) is correct.
Alternatively, we could have used the combination formula, which is the number of different ways of choosing k objects from n different objects. If nCk is the number of different subgroups of k objects that can be selected from a group of n different objects, then . Remember, in this formula,n ≥ k.
Now let’s plug numbers into the formula so that n is greater than or equal to k. We know we need 4 cards, so n = 4. Also, k is always less than or equal to n, and in this case we know we need 3 of the "low" cards, so k = 3. The number of possibilities is
This method also brings us to Answer Choice (D). The combination formula is the concrete math underlying this problem, but the key step here is realizing that it is unnecessary; it is much faster and easier just to count the possibilities."
However, I still have some confusion. I believe that this is a combination problem not a permutation one as the question basically asks us to choose cards (1 from the 26 High cards & 3 from the 26 Low cards). The arrangement of the cards is not moot. Therefore the answer should be:
26C1 x 26C3
Following is the solution from Kaplan which I somehow did not find very convincing. Although they have confirmed that this is a combination problem, they have applied the formula to only the 4 cards and not the selection from the entire batch of 52 cards:
"First, we need to figure out how many cards of each type we will need in order to earn 5 points. The "low" cards are worth 1 point. Since the question says we must draw 3 of these "low" cards, we get 3 points from "low" cards. This leaves us with 5 – 3 = 2 points left that we need to earn with the "high" cards. Since "high" cards are worth 2 points each, if we get one "high" card, we will reach our goal of 5 points total. Hence it takes us four cards to get 5 points: 3 "low" and 1 "high."
Our task is now simply to discover how many ways we can draw these 4 cards. Let’s draw out the possible ways: (L L L H), (L L H L), (L H L L), (H L L L) for a total of 4 different ways to get 3 "low" and 1 "high." Answer Choice (D) is correct.
Alternatively, we could have used the combination formula, which is the number of different ways of choosing k objects from n different objects. If nCk is the number of different subgroups of k objects that can be selected from a group of n different objects, then . Remember, in this formula,n ≥ k.
Now let’s plug numbers into the formula so that n is greater than or equal to k. We know we need 4 cards, so n = 4. Also, k is always less than or equal to n, and in this case we know we need 3 of the "low" cards, so k = 3. The number of possibilities is
This method also brings us to Answer Choice (D). The combination formula is the concrete math underlying this problem, but the key step here is realizing that it is unnecessary; it is much faster and easier just to count the possibilities."
