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# A fortune-teller uses 22 Major Arcana Tarot cards to perform

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A fortune-teller uses 22 Major Arcana Tarot cards to perform [#permalink]

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06 Oct 2008, 22:34
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A fortune-teller uses 22 Major Arcana Tarot cards to perform fortune telling. She will pick 3 cards to look at querent's past, present, and futuer. A card will have different meaning if it is reversed. How many possible ways can the fortune-teller read the fortune?

A) 1,540
B) 9,240
C) 12,320
D) 73,920
E) 79,464

Bonus: Same question. If she picks 10 cards, how many ways can she read the fortune?

Last edited by devilmirror on 07 Oct 2008, 09:06, edited 2 times in total.

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06 Oct 2008, 23:34
For the original question, the answer will be 44P3 = 79464.

If she picks 10 cards, the answer will be 44P10.

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07 Oct 2008, 09:07
Do you guys all agree with the above solution?

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08 Oct 2008, 23:05

I think E is right one.. here order will matter so we need to use P instead of C and reverse means we need to just double the card..

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10 Oct 2008, 00:24
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devilmirror wrote:
A fortune-teller uses 22 Major Arcana Tarot cards to perform fortune telling. She will pick 3 cards to look at querent's past, present, and futuer. A card will have different meaning if it is reversed. How many possible ways can the fortune-teller read the fortune?

A) 1,540
B) 9,240
C) 12,320
D) 73,920
E) 79,464

Bonus: Same question. If she picks 10 cards, how many ways can she read the fortune?

First card; 22 cards available and 2 possible interpretations = 22 x 2
Second card; 21 cards available and 2 possible interpretations = 21 x 2
Third card; 20 cards available and 2 possible interpretations = 20 x 2

Therefore, n(events) = (22x2)(21x2)(20x2) = 73,920

OR
n(events) = ($$2^3$$)(P(22,3) = (8)(22x21x20) = 73,920

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10 Oct 2008, 00:32
devilmirror wrote:
devilmirror wrote:
A fortune-teller uses 22 Major Arcana Tarot cards to perform fortune telling. She will pick 3 cards to look at querent's past, present, and futuer. A card will have different meaning if it is reversed. How many possible ways can the fortune-teller read the fortune?

A) 1,540
B) 9,240
C) 12,320
D) 73,920
E) 79,464

Bonus: Same question. If she picks 10 cards, how many ways can she read the fortune?

First card; 22 cards available and 2 possible interpretations = 22 x 2
Second card; 21 cards available and 2 possible interpretations = 21 x 2
Third card; 20 cards available and 2 possible interpretations = 20 x 2

Therefore, n(events) = (22x2)(21x2)(20x2) = 73,920

OR
n(events) = ($$2^3$$)(P(22,3) = (8)(22x21x20) = 73,920

I am puzzled. If three cards are used to tell past, present and future, will the order of these cards not matter?

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10 Oct 2008, 00:37
devilmirror wrote:

Bonus: Same question. If she picks 10 cards, how many ways can she read the fortune?

n(events) = (2^10)(P(22,10))
= (1024)( 22! / (22-10)! ) = (1024)(22! / 12!) = (1024)(22x21x20x...x13)

= 2,402,866,180,915,200
= 2.4 thousand trillions

I guess fortune tellers will never run out of their stories.

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10 Oct 2008, 00:45
scthakur wrote:
I am puzzled. If three cards are used to tell past, present and future, will the order of these cards not matter?

Because the order of card is matter. That's why we use permutation.

Permutation: P(22,3) = 22 x 21 x 20 = 9,240

Note:
Combination: C(22,3) = 22 x 21 x 20 / 3! = 22 x 7 x 10 = 1,540

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10 Oct 2008, 01:43
devilmirror wrote:
scthakur wrote:
I am puzzled. If three cards are used to tell past, present and future, will the order of these cards not matter?

Because the order of card is matter. That's why we use permutation.

Permutation: P(22,3) = 22 x 21 x 20 = 9,240

Note:
Combination: C(22,3) = 22 x 21 x 20 / 3! = 22 x 7 x 10 = 1,540

Then, where am I going wrong in my interpretation?
There are 22 cards and each card has two faces. Hence, there are a total of 44 cards to choose from. If I have to choose 3 out of 44 cards then the possible ways could be 44P3.

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10 Oct 2008, 02:09
scthakur wrote:
devilmirror wrote:
scthakur wrote:
I am puzzled. If three cards are used to tell past, present and future, will the order of these cards not matter?

Because the order of card is matter. That's why we use permutation.

Permutation: P(22,3) = 22 x 21 x 20 = 9,240

Note:
Combination: C(22,3) = 22 x 21 x 20 / 3! = 22 x 7 x 10 = 1,540

Then, where am I going wrong in my interpretation?
There are 22 cards and each card has two faces. Hence, there are a total of 44 cards to choose from. If I have to choose 3 out of 44 cards then the possible ways could be 44P3.

If you Assume number of cards is 22x2 = 44 , due to 2 faces than permutations of chossing a card out of a pack of 44 is different. Now you have only 22 cards to choose and each have 2 faces. Hope it clear your doubt !

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10 Oct 2008, 02:45
AmitGUPTA wrote:
If you Assume number of cards is 22x2 = 44 , due to 2 faces than permutations of chossing a card out of a pack of 44 is different. Now you have only 22 cards to choose and each have 2 faces. Hope it clear your doubt !

Thanks Amit. I know what mistake I made.

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10 Oct 2008, 03:10
Good work devilmirror ....

Just reminded me of concepts of permutation that is order is important in permutation.. and in this case order is important.
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Re: PS: Tarot Cards   [#permalink] 10 Oct 2008, 03:10
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