Maths permutation and probability

Thinzar · August 15, 2023, 3:49am

I need help on this one. Could someone help me?

Juliette · August 15, 2023, 6:11am

hi is the answer A)?

possible combinations
123 132 231 213 312 321 222 114 411 141
P(all have unique numbers) = 6/10

Thinzar · August 15, 2023, 9:12am

The answer is A.
So I have to write down all the possibilities. Is there any other way to solve this.