Problem:
Let be the set of permutations of the sequence for which the first term is not 1 . A permutation is chosen randomly from . The probability that the second term is 2 , in lowest terms, is . What is ?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Since the first term is not 1 , the probability that it is 2 is . If the first term is 2 , then the second term cannot be 2 . If the first term is not 2 , there are four equally likely values, including 2 , for the second term. Thus the probability that the second term is 2 is
so .
\section*{OR}
The set contains (4)(4!) = 96 permutations, since there are 4 choices for the first term, and for each of these choices there are 4 ! arrangements of the remaining terms. The number of permutations in whose second term is 2 is , since there are 3 choices for the first term, and for each of these choices there are 3 ! arrangements of the last 3 terms. Thus the requested probability is , and .
The problems on this page are the property of the MAA's American Mathematics Competitions