How many different combinations can be used for a combination lock with 60 numbers.

Question

1. How many different combinations can be used for a combination lock with 60 numbers.

(a) if it takes three numbers to unlock the lock?

(b) if the three numbers must be unique?

(c) if the consecutive numbers in the combination must be unique?

I'm confused with this question, please give an explanation + steps if you can, thank you.

Answer 1

I want to add my 2 cents to Michael M.'s answer.,

If there are 60 numbers to pick from and 3 numbers in the combination,

then in the first scenario, you can get 60³ ways because repeats are allowed.

In the second scenario, as Michael said, you can have 60*59*58 possible combinations,

If in scenario 3 you mean 3 specific numbers in a specific sequence, then only one way.

Answer 2

So there are 60 numbers. From the 60 numbers, you are assigning one to be the first number in the combination, one to be the second, one to be the third, and the other 57 are null.

First: 1

Second: 1

Third: 1

Null: 57

The different number of ways you can assign this is the total number factorial divided by each of the parts factorial

So: 60! / (1! * 1! * 1! * 57!) = ₆₀P₃ = 205320

For the third scenario, the first number can be any of the 60 numbers. The second number can't be the same as the first, so that means there are 59 possible ways of choosing the second number. Let's say the first number was 10. The second number can be any of the other 59 numbers. The third number can't be the same as the second number. Therefore, there are 59 ways of choosing the third number.

The total number of combinations for the third scenario, therefore is 60*59*59