2003 AMC 12B Problems/Problem 19
Problem
Let be the set of permutations of the sequence
for which the first term is not
. A permutation is chosen randomly from
. The probability that the second term is
, in lowest terms, is
. What is
?
Solution
There are choices for the first element of
, and for each of these choices there are
ways to arrange the remaining elements. If the second element must be
, then there are only
choices for the first element and
ways to arrange the remaining elements. Hence the answer is
, and
.
Solution 2
There is a chance that the number
is the second term. Let
be the chance that
will be the second term. Since
and
are in similar situations as
, this becomes
Solving for , we find it equals
, therefore
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=1215
~ pi_is_3.14
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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