Difference between revisions of "2020 AMC 8 Problems/Problem 10"
(→Solution 2 (complementary counting)) |
(→Solution 1) |
||
Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
− | First we need to count the the total number of possibilities. That is 4!=24. Now we subtract the cases where the Tiger and the Steelie are together. Doing this, we get 24-6= | + | First we need to count the the total number of possibilities. That is 4!=24. Now we subtract the cases where the Tiger and the Steelie are together. Doing this, we get 24-6=18 or D. |
==Solution 2 (complementary counting)== | ==Solution 2 (complementary counting)== |
Revision as of 10:42, 25 November 2020
Contents
Problem
Zara has a collection of marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
Solution 1
First we need to count the the total number of possibilities. That is 4!=24. Now we subtract the cases where the Tiger and the Steelie are together. Doing this, we get 24-6=18 or D.
Solution 2 (complementary counting)
There would be ways to arrange the
marbles, except for the condition that the Steelie and Tiger cannot be next to each other. If we did place them next to each other with the Steelie first, there would be
ways to place them (namely
,
; or
, where
and
denote the Steelie and the Tiger as in Solution 1). Accounting for the other possible order, there are a total of
ways. Now, there are
ways to place
and
, giving overall
ways to arrange the marbles with
and
next to each other. Subtracting this from
(to remove the cases which are not allowed) gives
valid ways to arrange the marbles.
Solution 3 (variant of Solution 2)
As in Solution 2, there are total ways to arrange the marbles without any constraints. To count the number of ways where the Steelie and the Tiger are next to each other, we treat them together as a "super marble". There are
ways to arrange the Steelie and Tiger within the super marble, then
ways to put the super marble in a row with the Aggie and the Bumblebee. Thus the answer is
.
Video Solution
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.