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The question was posed as: Marissa is doing a Tarot reading in which she must pick 6 cards from a deck of 72. The order of their selection is not important. Marissa does not want to see the Fool card. How many of the possible readings do not feature the Fool?

What I think I should do is find the total amount of possibilities of 6 cards out of 72, C(72,6)= n! r!(n-r)!

  =__72!__
     6!(72-6)!

  =__72!__
       6!66! 

  =__6.12e+103_
    (720)(5.44e+92)

  =6.12e+103
     3.92e+95

  =156,238,908 possible combination of cards

Then, find the amount out of a deck of 71 C(71,6) = 71! 6!65!

    =__8.50e+101__
     (720)(8.25e+90)

    =__8.50e+101__
       5.94e+93

    =143218999

Then subtract them 156238908-143218999= 13019909

That number is way too low, so that's where I am stuck.

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Is there only one fool card in the deck? –  Ataraxia Jun 21 '13 at 1:32
    
That would be correct –  Victoria Jun 21 '13 at 1:58
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1 Answer 1

up vote 2 down vote accepted

You're close. You need to find the number of 6 card hands that contain the fool card:

$$\binom{71}{5}$$

Then subtract that number from the total number of 6 card hands in the deck:

$$\binom{72}{6}-\binom{71}{5}$$

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Just so I understand the difference, instead of C(71,6) it would be C(71,5)? –  Victoria Jun 21 '13 at 1:59
    
@Victoria Yea, that's it. You had the right idea, just that one detail. –  Ataraxia Jun 21 '13 at 2:46
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