Difference between revisions of "2013 AIME II Problems/Problem 11"
(→Solution) |
(→Solution) |
||
| Line 13: | Line 13: | ||
Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>. | Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
==See Also== | ==See Also== | ||
Revision as of 14:20, 8 January 2023
Contents
Problem 11
Let
, and let
be the number of functions
from set
to set
such that
is a constant function. Find the remainder when
is divided by
.
Solution
Any such function can be constructed by distributing the elements of
on three tiers.
The bottom tier contains the constant value,
for any
. (Obviously
.)
The middle tier contains
elements
such that
, where
.
The top tier contains
elements such that
equals an element on the middle tier.
There are
choices for
. Then for a given
, there are
ways to choose the elements on the middle tier, and then
ways to draw arrows down from elements on the top tier to elements on the middle tier.
Thus
, giving the answer
.
Solution 2
See Also
| 2013 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.