To find the probability that a student chooses one art elective and one history elective, given three art electives, four history electives, and five computer electives, while each student can choose two electives, we need to use combinatorial methods.
First, let's consider how many ways there are to choose one art elective out of three:
[tex]\[
{}_{3}C_1
\][/tex]
Next, we determine how many ways there are to choose one history elective out of four:
[tex]\[
{}_{4}C_1
\][/tex]
Now, we need to determine the total number of ways to choose 2 electives out of the available 12 electives (3 art + 4 history + 5 computer):
[tex]\[
{}_{12}C_2
\][/tex]
To find the probability that a student chooses one art elective and one history elective, we compute the product of the ways to choose these individual electives and divide by the total number of ways to choose any 2 electives from the 12:
[tex]\[
\frac{({}_{3}C_1) \times ({}_{4}C_1)}{{}_{12}C_2}
\][/tex]
The probability expression that represents this situation matches:
[tex]\[
\frac{\left({}_{3}C_1 \right) \left({}_{4}C_1 \right) }{{}_{12}C_2}
\][/tex]
Thus, the correct expression is:
[tex]\[
\frac{\left({}_{3}C_1 \right) \left({}_{4}C_1 \right) }{{}_{12}C_2}
\][/tex]
