Question

Alex must choose a 4-digit code for a PIN number from the digits 1 through 9,
without repeating any digits. What is the probability that the selected code:
a) is “1234”
b) starts with a “1”
c) contains a “1” in any position

Answer 1

How many permutations can you make with 9 digits, when you choose 4 of them?

Answer 2

126?

Answer 3

Use the formula:
$$\frac{n!}{(n-r)!(r!)}$$
And you get 126.

Answer 4

a) 1 / (9 * 8 * 7 * 6)
b) (1 * 8 * 7 * 6) / (9 * 8 * 7 * 6)
c) {(9 * 8 * 7 * 6) - (8 * 7 * 6 * 5)} / (9 * 8 * 7 * 6)

Answer 5

Now, 1234 is only one of these. So the probability is 1/126.

Answer 6

@bloopman, order matter in this case. (9P4) is 3024

Answer 7

You're right. I'm sorry.

Answer 8

Yes, Order matters so Permutations, not Combinations.

Answer 9

Thanks a lot sourwing and bloopman also.

Answer 10

for part c, do you meant EXACTLY ONE 1 or AT LEAST? My answer is for at least

Answer 11

I think it's 1 in any position in x x x x. And no repetitions. So Exactly one 1

Answer 12

ah that's right. but my answer still holds XD

Answer 13

Yeah I thought so since you made the numbers less by 1. Thanks alot buddy.