document.write( "Question 480841: A restaurant has 10 choices for appetizers, 20 choices for the main course, and 12 choices for the dessert. Of the 12 desserts half of them are sugar-free.
\n" ); document.write( "What is the probability that a person chooses 3 appetizers, 2 main course items, and 2 desserts in such a way that exactly both are sugar-free? \n" ); document.write( "
Algebra.Com's Answer #329242 by Theo(13342)\"\" \"About 
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without worrying about sugar-free desserts, the number of possible combinations are:
\n" ); document.write( "10C3 * 20C2 * 12C2
\n" ); document.write( "with worrying about sugar-free desserts, the number of possible combinations are:
\n" ); document.write( "10C3 * 20C2 * 6C2
\n" ); document.write( "the probability will be:
\n" ); document.write( "(10C3 * 20C2 * 6C2) / (10C3 * 20C2 * 12C2)
\n" ); document.write( "the 10C3 and the 20C2 in the numerator and the denominator cancel out and you are left with:
\n" ); document.write( "6C2 / 12C2
\n" ); document.write( "this becomes 15/66 which is equal to .227272727
\n" ); document.write( "that's the probability that you will get 2 desserts that are sugar-free with your meal.
\n" ); document.write( "this assumes that the order in which you receive the appetizers and the main courses and the desserts is not important.
\n" ); document.write( "if order is important, then you would use the permutations formula to get:
\n" ); document.write( "6P2 / 12P2 = .227272727
\n" ); document.write( "the probability is the same.
\n" ); document.write( "6C2 = 15
\n" ); document.write( "12C2 = 66
\n" ); document.write( "6P2 = 30
\n" ); document.write( "12P2 = 132
\n" ); document.write( "15/66 is the same ratio as 30/132
\n" ); document.write( "the combination formula is:
\n" ); document.write( "nCx = n! / ((n-x)! * x!)
\n" ); document.write( "the permutation formula is:
\n" ); document.write( "nPx = n! / (n-x)!
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