I think I got this. I can't think of a less difficult way.
To get at least 14 questions right, Beavis has to get 14, 15, 16, 17, 18 or 20 questions right. He can't get 19 questions right because getting 19 of 20 right means the 20th question must be right.
There are 20! different ways to arrange the 20 unique answers. 20!=20*19*18*...*3*2*1=a huge number, approximately 2.4 million times a trillion. Only one of those ways results in getting all 20 questions right.
To get 18 right and 2 wrong, two of the answers must be switched (e.g. switch the correct answers for #7 and #15 and he will get two wrong). There are 20 choose 2 ways to do this. 20 choose 2 is 20!/(18!*2!) = 20*19/2 = 190 ways.
Next, to get 17 right and 3 wrong, three answers must be switched around to be incorrect. This is where it starts to get difficult. Let's say that we have correct answers 1,2,3. There are 3!=6 ways to arrange 1,2,3 (i.e. 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,1,2; 3,2,1). But of those 6 ways, 4 of them contain correct answers. 1,2,3 is all correct. 1,3,2 is correct for 1. 2,1,3 is correct for 3. 3,2,1 is correct for 2. Only 2,3,1 and 3,1,2 are all wrong.
There are 20 choose 3 ways to pick 3 of 20 unique items, each of which has 2 ways within it to arrange the wrong answers. So we have 20!/(17!*3!) * 2 = 1140 ways to get 17 right and 3 wrong.
To get 16 right and 4 wrong, four right answers must be switched around and all four stay wrong. Rather than count them, I figured it out this way:
There are 4! ways to arrange 4 unique answers. Of the 4!=24 ways, 1 is all four correct. Getting 3 correct and 1 wrong is not possible, because if you got 3 correct of 4, then you've got all 4. Getting 2 correct and 2 wrong has 4 choose 2 = 4!/(2!*2!) = 6 ways. Of the 2 wrong ways, there is only one way to arrange them. Finally, to get 1 correct and 3 wrong, there are 4 choices for right and for each one, 2 ways to arrange the other 3 wrong answers (total 4*2=8). This totals 1+6+8=15. Of all 24 ways to arrange 4 unique answers, 15 of them include at least one correct answer, leaving us with 24-15=9 ways to arrange four unique answers that are all wrong.
I manually counted 24 ways and identified 9 to make sure this is correct. At first I got 10 and then I found a mistake, so it is 9.
There are 20 choose 4 ways to pick 4 unique items of 20 (i.e. the four wrong answers), each of which has 9 different ways to arrange them, for a total of 20!/(16!*4!) * 9 = 43,605.
For 15 right and 5 wrong, the same calculation goes: 20 choose 5 = 15,504. Of the 5 answers we want to be incorrect, there are 5!=120 ways to arrange 5 unique items, but some of those will be correct which we do not want to count. Of those 120, 1 is all five correct. 4 of 5 is not possible. 3 right and 2 wrong is 5 choose 3 = 10. 2 right and 3 wrong is 5 choose 2 times 2 (the second 2 being how many ways can you arrange 3 wrong answers), which is 20. 1 right is 5 ways, each of which has 9 ways to arrange the 4 wrong answers, which is 5*9=45 (we just calculated the 9 above). The total is 1+10+20+45=76. 120-76=44. For each of the 20 choose 5 = 15,504, there are 44 ways to arrange the 5 to be wrong, for a total of 15504*44=682,176.
Lastly, we have 14 right and 6 wrong. In my Excel sheet, I typed in the formula =FACT(6)-1-FACT(6)/(FACT(4)*FACT(2))-FACT(6)/(FACT(3)*FACT(3))*2-FACT(6)/(FACT(4)*FACT(2))*9-6*44
which is 265
20 choose 6 times 265 is 10271400.
The total number of ways to arrange 20 unique answers and get 0 wrong, 2 wrong, 3 wrong, 4 wrong, 5 wrong, or 6 wrong is 1+190+2280+43605+682176+10271400=10999652.
While this number may seem big, it's a tiny fraction of 20 factorial. 10999652/20! = 4.52E-12, or 1 in 221 billion. Wow!
