Part A
There are 6 bijections from {1,2,3} to {a,b,c}. This is effectively the same as asking the question "how many ways are there to arrange {a,b,c} where order matters?" We use a factorial to answer this question.
3 factorial = 3! = 3*2*1 = 6
You can also use a permutation, which is composed of factorials, to get the same answer.
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Part B
There are no bijections from {1,2,3} to {a,b,c,d}. Why is this? Because a bijection has two properties: it must be one-to-one, and it must be onto. The term "onto" in mathematics means "every value in the range is targeted". In the case of the range {a,b,c,d} it is not possible for each value to show up. This is because there are only three items in the domain {1,2,3}. You'll always be one letter short.
As you can probably guess, a bijection is only possible if and only if n(D) = n(R), where D and R are the domain and range respectively. The notation n(D) represents the count or number of items in set D.
