Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 6"
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'''Method 2:''' Each element of <math>\mathcal{S}</math> has <math>3</math> options: be in neither <math>A</math> nor <math>B</math>, be in just <math>B</math>, or be in both <math>A</math> and <math>B</math>. All elements assort independently, so there are <math>3^{10}</math> valid pairs of subsets. | '''Method 2:''' Each element of <math>\mathcal{S}</math> has <math>3</math> options: be in neither <math>A</math> nor <math>B</math>, be in just <math>B</math>, or be in both <math>A</math> and <math>B</math>. All elements assort independently, so there are <math>3^{10}</math> valid pairs of subsets. | ||
− | Then the answer is <math>\frac{3^10}{2^20} \Rightarrow \boxed{205}.</math> | + | Then the answer is <math>\frac{3^{10}}{2^{20}} \Rightarrow \boxed{205}.</math> |
Latest revision as of 14:20, 9 August 2021
Problem 6
Let . Two subsets,
and
, of
are chosen randomly with replacement, with
chosen after
. The probability that
is a subset of
can be written as
, for some primes
and
. Find
.
Solution
Claim: The number of pairs and
such that
is
.
Method 1: If has
elements, then there are
choices for
. Since
has
elements
times as
ranges over all possible subsets of
, the desired quantity is
Method 2: Each element of has
options: be in neither
nor
, be in just
, or be in both
and
. All elements assort independently, so there are
valid pairs of subsets.
Then the answer is