Question 17 of 18, Step 1 of 1 $14 / 18$ 2 Correct Incorrect A university is putting together a hiring commitcee of 6 members for the next university provost. The eligible pool of members consirts of 12 administrative personnel, 5 faculty, and 7 students in how many ways can the committee be formed if there must be 2 members chosen from each of the three subgroups? Answer Keypad Kejbourd Shortcuts

Step 1 ：The problem is asking for the number of ways a committee can be formed with 2 members from each of the three subgroups: administrative personnel, faculty, and students. This is a combination problem because the order in which the members are chosen does not matter.

Step 2 ：The formula for combination is \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to choose, and ! denotes factorial.

Step 3 ：We need to calculate the combination for each subgroup and then multiply them together to get the total number of ways the committee can be formed.

Step 4 ：For the administrative personnel subgroup, there are 12 members and we need to choose 2. Using the combination formula, we get \(C(12, 2) = \frac{12!}{2!(12-2)!} = 66\).

Step 5 ：For the faculty subgroup, there are 5 members and we need to choose 2. Using the combination formula, we get \(C(5, 2) = \frac{5!}{2!(5-2)!} = 10\).

Step 6 ：For the students subgroup, there are 7 members and we need to choose 2. Using the combination formula, we get \(C(7, 2) = \frac{7!}{2!(7-2)!} = 21\).

Step 7 ：To get the total number of ways the committee can be formed, we multiply the combinations of each subgroup together. So, \(66 * 10 * 21 = 13860\).

Step 8 ：Final Answer: The total number of ways the committee can be formed is \(\boxed{13860}\).