<PRE><B>Question 188411</B>
Art, Becky, Carl, Denise are lined up to buy tickets. 
A. How many different permutations are possible?
Ans: 4! = 4*3*2*1 = 24
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B. Suppose Ed was also in line. How many permutations would there be?
Ans: 5! = 5*4*3*2*! = 120
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C. In how many of the permutations of the five is Becky first?
Ans: 1*4! = 24
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D. What is the probability that a permutation of this five is chosen at random will have Becky first?
Ans: 24/120 = 0.2
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Cheers,
Stan H.
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