Write in set builder form {all possible numbers formed by any two of the digits 1 2 5}
With 3 numbers, we got 3! = 6 possible numbers formed by the two digits
{x : x ∈ {12, 15, 21, 25, 51, 52})
x such that x is a element of the set {12, 15, 21, 25, 51, 52}
With 3 numbers, we got 3! = 6 possible numbers formed by the two digits
- 12
- 15
- 21
- 25
- 51
- 52
{x : x ∈ {12, 15, 21, 25, 51, 52})
x such that x is a element of the set {12, 15, 21, 25, 51, 52}