How many ways can 8 people stand in a line if Alice and Bob refuse to stand next to each other?

Explanation:

We have 8 people to start with. If we remove Alice and Bob, and replace them with Charlie (who will be a stand in for both people), then we have 8-2+1 = 7 people in this line. There are 7! = 7*6*5*4*3*2*1 = 5040 different permutations or line orderings for these seven people.

For any given permutation, replace Charlie with Alice and Bob. There are two ways to do this for any ordering. We could have Alice in front of Bob, or Bob in front of Alice. So there are 2 times as many permutations compared to 5040. In other words, there are 2*5040 = 10,080 different permutations where Alice and Bob are standing together.

This is out of 8! = 8*7*6*5*4*3*2*1 = 40,320 different permutations overall of arranging 8 people in a line.

This means there are 40,320 - 10,080 = 30,240 different ways to arrange 8 people such that Alice and Bob are not standing together.

In summary, the idea is to find out how many ways there are to have Alice and Bob together. Then we subtract that result from the total number of ways to arrange 8 people to get our final answer.

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.

idk

step-by-step explanation: