Question 188411: Art, Becky, Carl, Denise are lined up to buy tickets.
A. How many different permutations are possible?
B. Suppose Ed was also in line. How many permutations would there be?
C. In how many of the permutations of the five is Becky first?
D. What is the probability that a permutation of this five is chosen at random will have Becky first?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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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