A. determine all bijections from the {1,2,3} into {a, b,c}. b. determine all bijections from {1, 2, 3} into {a, b,c, d}.

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.

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.

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.

yes, whats the problem?