Difference between revisions of "2021 AMC 12A Problems/Problem 15"
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<li>If <math>t-b=4,</math> then <math>\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}</math> different groups can be formed.</li><p> | <li>If <math>t-b=4,</math> then <math>\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}</math> different groups can be formed.</li><p> | ||
</ol> | </ol> | ||
− | By the combinatorial identity <math>\binom{n}{k}=\binom{n | + | By the combinatorial identity <math>\binom{n}{k}=\binom{n}{n-k}</math> and Vandermonde's Identity <math>\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r},</math> we find the total number of such groups: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
N&=\binom{6}{0}\binom{8}{8}+\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{k+4}\right]+\left[\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{k}\right]-1\right]+\left[\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}\right] \\ | N&=\binom{6}{0}\binom{8}{8}+\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{k+4}\right]+\left[\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{k}\right]-1\right]+\left[\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}\right] \\ | ||
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from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math> | from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math> | ||
− | + | Alternatively, since the answer choices have different units digits, it suffices to find the units digit of <math>N</math> only. | |
− | ~MRENTHUSIASM | + | ~sugar_rush ~MRENTHUSIASM |
+ | |||
+ | ==Solution 5 (Symmetry Applied Twice)== | ||
+ | |||
+ | Consider the set of all <math>2^{8+6}=2^{14}</math> possible choirs that can be formed. For a given choir let <math>D</math> be the difference in the number of tenors and bases modulo <math>4</math>, so <math>D = T - B \pmod{4}.</math> Exactly half of all choirs have either <math>D=0</math> or <math>D=2</math>. To see this, pick one of the tenors and note that including or removing him from a choir changes <math>D</math> by <math>\pm1</math>. Of those <math>2^{13}</math> choirs with <math>D=0</math> or <math>D=2</math>, we claim exactly half have <math>D=0</math>. To see this, for any choir having <math>D=0</math> or <math>D=2</math>, we can replace the <math>T</math> tenors with the <math>6 - T</math> tenors who were not in the choir, thereby sending <math>D \mapsto D + 2 \pmod{4}.</math> Excluding the empty choir, there are <math>2^{12}-1 = 4095</math> choirs that meet the conditions of the problem, and the answer is <math>\boxed{\textbf{(D) } 95}</math>. | ||
− | + | ~telluridetoaster and ~bigskystomper | |
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Latest revision as of 05:40, 26 July 2024
Contents
- 1 Problem
- 2 Solution 1 (Bijection)
- 3 Solution 2 (Vandermonde's Identity)
- 4 Solution 3 (Generating Functions)
- 5 Solution 4 (Enumeration)
- 6 Solution 5 (Symmetry Applied Twice)
- 7 Video Solution by Punxsutawney Phil
- 8 Video Solution by Hawk Math
- 9 Video Solution by OmegaLearn (Using Vandermonde's Identity)
- 10 See also
Problem
A choir director 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 (Bijection)
Suppose that tenors and
basses are selected. The requirements are
and
It follows that basses are not selected. Since the ordered pairs
and the ordered pairs
have one-to-one correspondence, we consider the ordered pairs
instead. The requirements become
and
which simplify to
and
respectively.
As the total number of such groups is
from which
~MRENTHUSIASM
Solution 2 (Vandermonde's Identity)
Suppose that tenors and
basses are selected. The requirements are
and
Note that different groups can be formed by selecting
tenors and
basses. Since
we apply casework:
- If
then
different group can be formed.
- If
then
different groups can be formed.
- If
then
different groups can be formed, recalling that
- If
then
different groups can be formed.
By the combinatorial identity and Vandermonde's Identity
we find the total number of such groups:
from which
~MRENTHUSIASM
Solution 3 (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
, then
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
.
~lawliet163
Solution 4 (Enumeration)
Note that different groups can be formed by selecting
tenors and
basses. By casework, we construct the following table:
We find the total number of such groups:
from which
Alternatively, since the answer choices have different units digits, it suffices to find the units digit of only.
~sugar_rush ~MRENTHUSIASM
Solution 5 (Symmetry Applied Twice)
Consider the set of all possible choirs that can be formed. For a given choir let
be the difference in the number of tenors and bases modulo
, so
Exactly half of all choirs have either
or
. To see this, pick one of the tenors and note that including or removing him from a choir changes
by
. Of those
choirs with
or
, we claim exactly half have
. To see this, for any choir having
or
, we can replace the
tenors with the
tenors who were not in the choir, thereby sending
Excluding the empty choir, there are
choirs that meet the conditions of the problem, and the answer is
.
~telluridetoaster and ~bigskystomper
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.