Difference between revisions of "2024 SSMO Relay Round 4 Problems/Problem 1"
(Created page with "==Problem== Freddy the Frog can jump <math>1</math> unit right or up. He is at <math>(1,1)</math> and wants to get to <math>(7, 4)</math>. However, he is scared of points <ma...") |
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==Solution== | ==Solution== | ||
| + | Using the Principle of Inclusion and Exclusion, the answer is the number of ways to get to <math>(7,4)</math> minus the number of ways to get to <math>(7,4)</math> passing through <math>(3,1)</math> minus the number of ways to get to <math>(7,4)</math> passing through <math>(3,2)</math> plus the number of ways to get to <math>(7,4)</math> passing through both <math>(3,1)</math> r <math>(3,2).</math> This is equivalent to <cmath>\binom{11}{4}-\binom{4}{1}\binom{7}{3}-\binom{5}{2}\binom{6}{2}+\binom{4}{1}\binom{6}{2} = 330-140-150+60 = \boxed{100}.</cmath> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:47, 10 September 2025
Problem
Freddy the Frog can jump 
unit right or up. He is at 
and wants to get to 
. However, he is scared of points 
and 
and will not hop onto those points. How many ways can he reach his destination?
Solution
Using the Principle of Inclusion and Exclusion, the answer is the number of ways to get to 
minus the number of ways to get to passing through 
minus the number of ways to get to passing through 
plus the number of ways to get to passing through both r 
This is equivalent to ![\[\binom{11}{4}-\binom{4}{1}\binom{7}{3}-\binom{5}{2}\binom{6}{2}+\binom{4}{1}\binom{6}{2} = 330-140-150+60 = \boxed{100}.\]](http://latex.artofproblemsolving.com/8/c/b/8cb5044161bc495aecabddc42d9754648c0ff34c.png)
~SMO_Team
