5
$\begingroup$

Timmy is preparing his math GRE subject test. Unfortunately Timmy sucks at math, but he has been practicing with a bunch of previous GRE tests. The GRE consists of $66$ question, since Timmy is no coward he answers every question with one of the options $a,b,c,d$ or $e$. Timmy knows that he needs $50$ questions to get into his desired college.

Since Timmy really sucks at math he is going to select the answers to the questions beforehand, even though he doesn't even know what the questions are!

All that Timmy knows is that the answer-code for the GRE never repeats, and Timmy has already taken $N$ previous GREs ( the answer-codes to these are random). What is the minimum number $N$ such that Timmy can always find a new answer-code that is sure to reach at least $50$ correct answers?

  • 1
    Timmy is gonna have to look through a lot of tests, I can tell you that much. – 2017-01-24

1 Answers 1

2

There are $4^{66-i}\binom{66}{i}$ answer keys that will end up giving Timmy $i$ correct answers - $\binom{66}{i}$ ways to choose the $i$ questions that will be answered correctly and $4^{66-i}$ ways to choose the answers for the incorrect questions. Since Timmy needs at least $50$ correct answers, the total number of possible answer keys is $$\sum_{i=50}^{66} 4^{66-i}\binom{66}{i}=984401002589851920172825$$ (as computed by Wolfram Alpha). Therefore, since there are a total of $5^{66}$ answer keys, Timmy will need to do at least $$5^{66}-984401002589851920172825=13552527156068805425092175609871681540902092800$$ answer keys.