2023 WSMO Accuracy Round Problems/Problem 2: Difference between revisions
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==Solution== | ==Solution== | ||
There are <math>6!</math> permuations of "precal", meaning it takes Bob <cmath>6!\cdot5 = 720\cdot5 = 3600\text{ seconds} \implies\frac{3600}{60} = 60\text{ minutes}</cmath> to write all the permutations. There are <math>\tfrac{8!}{2!2!2!}=5040</math> permutations of "calculus", meaning it takes Bob <cmath>5040\cdot2 = 10080\text{ seconds}\implies \frac{10080}{60} = 168\text{ minutes}</cmath> to write all the permutations. Thus, our answer is <cmath>168-60 = \boxed{108}.</cmath> | |||
~pinkpig | |||
Latest revision as of 10:41, 13 September 2025
Problem
When Bob is in precalculus, he gets bored and writes all the permutations in "precal". Since he is not very smart, it takes him 5 seconds to write each permutation. When Bob advances to calculus, he gets bored and writes all the permutations in "calculus". He is smart and can now write each permutation in 2 seconds. Find the positive difference in minutes between the time it takes for him to write the permutations of "precal" and "calculus".
Solution
There are 
permuations of "precal", meaning it takes Bob ![\[6!\cdot5 = 720\cdot5 = 3600\text{ seconds} \implies\frac{3600}{60} = 60\text{ minutes}\]](http://latex.artofproblemsolving.com/6/3/9/63923b47464b51401b0119094543ba3597adedd7.png)
to write all the permutations. There are 
permutations of "calculus", meaning it takes Bob ![\[5040\cdot2 = 10080\text{ seconds}\implies \frac{10080}{60} = 168\text{ minutes}\]](http://latex.artofproblemsolving.com/2/a/a/2aa558742b85a722a96cddf3cd7402daa7b373a2.png)
to write all the permutations. Thus, our answer is ![\[168-60 = \boxed{108}.\]](http://latex.artofproblemsolving.com/5/f/3/5f36a10cb4da9a501a5d0a5a8181a6923c104877.png)
~pinkpig
