Don't know if it is a trick question. My answer is 16/36.Ok for problem 8:

there are 36 possible outcomes

e isn't a trick question is it? it's basically the same as a thru d correct?

I agree.Problem 9: should be number of multiples of 7 plus multiples of 3 for 5 digits minus number of multiples for 21.

I draw a binary tree in case A wins the first game. From there I got 10 possible developments. There is the same number of developments when B wins the first game.Problem 10 is what I'm having trouble with. Seems like there are many ways total and I was going to use the probability tree but it doesn't work as well as it would on problem 11.

8 variants for the first character, 10 variants for each of the rest, so 8 * 10^4 total.Problem 11: used tree, got 88888 different pins.

It makes no sense why from P(n,r) = n(n-1)(n-2)...(n-r+1) and r = 5 you concluded P(n,5) = (n-5+1). Where are the other four factors? Also, n - 5 + 1 is not n - 6.14. I used a formula. P(n,r) = n(n-1)(n-2)...(n-r+1)

i set the equation as P(n,5) = (n-5+1) then n-6=2027400

got n as 2027406 and used that for (2027406)!/(5!*2027401!)

Note, however, that .