The planning committee at school has 11 members. Exactly five of these members are teachers. A four-person subcommittee with at least one member who is not a teacher must be formed from the members of the planning committee. How many distinct subcommittees are possible?

So the answer is 325

Because there are 5 teachers on the committee, there are 6 non-teachers. Now, in total, we can form (11 \choose 4) = 330 subcomittees. The number of subcommittees with zero non-teachers is the number of subcommittees formed by the 5 teachers, totaling (5 \choose 4) = 5. So, the number of subcomittees with at least one non-teacher is 330 - 5 = 325

Well, organizing a subcommittee can be quite a circus! Let's break it down. Since we must have at least one member who is not a teacher, we can choose one non-teacher from the six available. That gives us 6 choices for one member.

For the remaining three members, we can choose from the entire pool of 11 members, including both teachers and non-teachers. That gives us 11 choices for the second member, 11 choices for the third member, and 11 choices for the fourth member.

To find the total number of distinct subcommittees, we multiply the number of choices for each member together: 6 choices for the first member, multiplied by 11 choices for the second, multiplied by 11 choices for the third, and multiplied by 11 choices for the fourth.

Therefore, the total number of distinct subcommittees is 6 x 11 x 11 x 11 = 8,026.

So, it looks like there are 8,026 different ways to create this subcommittee. That's more options than a clown has colorful wigs!

To solve this problem, we can use the concept of combinations.

First, let's consider the different cases for the subcommittee composition:
Case 1: All four members are teachers
Case 2: Three members are teachers, and one is not
Case 3: Two members are teachers, and two are not

For Case 1, there are ${5 \choose 4}$ ways to select the four teachers from the five available. Since there is only one way to select all four teachers, the number of possible subcommittees for Case 1 is 1.

For Case 2, there are ${5 \choose 3}$ ways to choose three teachers from the five available and ${6 \choose 1}$ ways to select one non-teacher from the remaining six members. Multiplying these two values together, we get ${5 \choose 3} \cdot {6 \choose 1} = 5 \cdot 6 = 30$ possible subcommittees for Case 2.

For Case 3, there are ${5 \choose 2}$ ways to select two teachers from the five available and ${6 \choose 2}$ ways to choose two non-teachers from the remaining six members. Multiplying these two values together, we get ${5 \choose 2} \cdot {6 \choose 2} = 10 \cdot 15 = 150$ possible subcommittees for Case 3.

Adding up the number of subcommittees for each case, we have:
Total number of subcommittees = Number of subcommittees for Case 1 + Number of subcommittees for Case 2 + Number of subcommittees for Case 3
Total number of subcommittees = 1 + 30 + 150 = 181

Therefore, there are 181 distinct subcommittees possible with the given conditions.

so you have 5 teachers and 6 non-teachers

the cases you DON'T want is the case where all 4 that are selected are teachers
number of selections with no restrictions = C(11,4)
number of selections with all teachers = C(5,4)

so number of selections as stated = ....

You guys copied this off alcumus. Stop cheating.

its not gud 2 give answers w/out explaining