In a certain card game, an ace counts one point, face cards

Question

In a certain card game, an ace counts one point, face cards count 10 points, and all other cards count their face values. A player must play at least one card on each turn, and his score on a turn is determined by the total point value of the cards played. A player holds the following cards: ace of hearts, two of clubs, four of diamonds, 6 of hearts and 8 of spades. How many different point totals are possible for him on his next turn?

A) 31
B) 26
C) 21
D) 10
E) 23

Bluebird · Answer

I got A and here's how:

If the player decides to play only 1 card, he has 5 choices: 5
If the player decides to put down 2 cards he has 10 combinations to choose from (5!/2!3!=10): 10
If the player decides to put down 3 cards he has 10 combinations to choose from (5!/3!2!=10): 10
If the player decides to put down 4 cards he has 5 combinations to choose from (5!/4!1!=5): 5
If the player decides to put down all 5 cards he has only 1 choice.

Add all the possible choices together and you get: 5+10+10+5+1=31

Answer: A

ethan_casta · Answer

The answer above is incorrect. This is the case because there they are double counting some card hands. The point system advanced in the question results in at least 9 double counts. So, you have to subtract at least 9 from the previous respondants end result of 31. For example, a hand of 2, 4, and 6 equals 12 by the system of counting and a hand of 4, and 8 equals 12 also by the system of counting.

KillerSquirrel · Answer

I'm sorry, but i think your logic is incorrect

as was said before me: C(5,1) = 5 C(5,2) = 10 C(5,3) = 10 C(5,4) = 5 C(5,5) = 1 Total = 31 as for subtracting duplicates the combinations formula take care of that - see an example: 1,2,4,6,8 (cards in the player hand) C(5,2) = 10 1,2 1,4 1,6 1,8 2,4 2,6 2,8 4,6 4,8 6,8 Total = 10 :-D

ethan_casta · Answer

I think I'm right. In 5C2, for example, you are counting 8,4 as a unique hand, right. This sums to 12. In 5C3, you are counting 2,4,6, as a unique hand also. This sums to 12. So, when you add 5C2 to 5C3, you get two hands that both sum to 12. That should count as one hand not two because the end result is 12 in both cases. If that is incorrect, please explain how. Thanks.

ethan_casta · Answer

Note: the questions states "how many different point totals" in the last sentence.

KillerSquirrel · Answer

I see your point :-D

It's not realy about combinations , the only possible values are: 1,2,4,6,8 1 = 1 (one card of ace) 2 = 2 (one card of two) 1,2 = 3 (one card of ace, and one card of two) 4 = 4 4,1 = 5 2,4 = 6 6,1 = 7 8 = 8 8,1 = 9 and so on until finally reaching to: 1,2,4,6,8 = 21 so different point totals are 21 (1 through 21) and the answer is (C) excellent ethan_casta ! well done ! :-D

Caas · Answer

The OA is 21.

1) total dif points totals:
C(5,5) +C(5,4)+C(5,3)+C(5,2)+C(5,1)= 1+5+10+10+5=31
2) Points: 1 2 4 6 8
First consider only 2, 4,6 and 8 (without 1)
2+4=6
2+6=8
10=2=8, but also 4+6
12=8+4 but also 4+6+2
14=8+2+4 but also 8+6
So, there are 5 different pairs with equal totals (without 1)
Then add 1: it another 5 equal totals
Total will be 10 equal totals. 

3) Answer will be 31-10=21

ethan_casta · Answer

Thanks for helping to illustrate my previous posts Caas! Your proof is good and easy to follow.

In a certain card game, an ace counts one point, face cards

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