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Christmas Math Fun
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À
11/12/1997 17:53:33
Matt Mc Donnell
Mc Donnell Software Consulting
Boston, Massachusetts, États-Unis
Information générale
Forum:
Visual FoxPro
Catégorie:
Autre
Divers
Thread ID:
00064933
Message ID:
00065494
Vues:
50
>>>>>>>Types of problems like this fall into the category of cardinality. Some can't be solved without knowing the theory, since there is no "brute force" method. For example: You and your most significant other decide that for simplicity's sake you're only going to exchange 5 gifts, from a list of 52 (one for each week). Come Christmas morning, how many distinct possibilities (including opening order) are there?
>>>>>>>
>>>>>>You're right, I wouldn't want to try solving this without a knowledge of permutations...
>>>>>
>>>>>A bunch !!!!!
>>>>
>>>>Did I express the problem wrong? It doesn't involve permutations, but rather distinct combinations. Perhaps I should have left the "opening order" business out.
>>>>
>>>Ah, then we're back to combinations...
>>
>>Yep!
>
>This looks simple enough to me:
>
>number of gift combinations * number of opening orders=
>(52 * 51 * 50 * 49 * 48 * 47 * 46 * 45 * 44 * 43) * ;
>(10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2) = ;
>2.08321075874349 * 10^23

Ah Matt, is this the same as 52! / 47!? I don't think so. The result I get for that is:

2.586232415112e+59

Actually, I was trying for combinations and should have left out opening order.

George
George

Ubi caritas et amor, deus ibi est
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