Difference between revisions of "2021 AMC 12A Problems/Problem 15"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution | + | ==Solution 4 (Combinatoric Argument)== |
<b>We claim that if the empty group is allowed, then there are <math>\mathbf{2^{12}}</math> ways to choose the singers satisfying the requirements.</b> | <b>We claim that if the empty group is allowed, then there are <math>\mathbf{2^{12}}</math> ways to choose the singers satisfying the requirements.</b> | ||
Revision as of 02:59, 17 February 2021
Contents
Problem
A choir direction must select a group of singers from among his tenors and
basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of
, and the group must have at least one singer. Let
be the number of different groups that could be selected. What is the remainder when
is divided by
?
Solution 1
We know the choose function and we know the pair multiplication so we do the multiplications and additions.
~Lopkiloinm
Solution 2 (Generating Functions)
The problem can be done using a roots of unity filter. Let . By expanding the binomials and distributing,
is the generating function for different groups of basses and tenors. That is,
where
is the number of groups of
basses and
tenors. What we want to do is sum up all values of
for which
except for
. To do this, define a new function
Now we just need to sum all coefficients of
for which
. Consider a monomial
. If
,
otherwise,
is a sum of these monomials so this gives us a method to determine the sum we're looking for:
(since
and it can be checked that
). Hence, the answer is
with the
for
which gives
.
~lawliet163
Solution 3 (Casework and Vandermonde's Identity)
We will use the Vandermonde's Identity to find the requested sum:
~MRENTHUSIASM
Solution 4 (Combinatoric Argument)
We claim that if the empty group is allowed, then there are ways to choose the singers satisfying the requirements.
First, we set one tenor and one bass aside. We argue that each group from the remaining singers (of any size, including
) corresponds to exactly one desired group from the original
singers.
The remaining singers can form
groups. The left side counts directly, while the right side uses casework (selecting
tenors and
basses for each group). Now, we map each group from the
to a group from the
By casework:
Clearly, the mapping is satisfied. For each group from the we can obtain a desired group from the
by adding one tenor or one bass accordingly.
Since we can select
basses instead of
basses, without changing the number of groups. Therefore, we have the absolute difference
Since
we conclude that
and the mapping is satisfied by case
By the same reasoning as Case we select
basses instead of
basses. The absolute difference also is
Since
is even, it follows that
is also even, and
The mapping is satisfied by case
The proof of the bolded claim is complete.
Therefore, excluding the empty group gives
~MRENTHUSIASM
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=FD9BE7hpRvg&t=533s
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (using Vandermonde's Identity)
https://www.youtube.com/watch?v=mki7xtZLk1I
~pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.