How many people does it take to form a line?

A group of 10 people needs to form a line. The line will consist of exactly 8 of the people. Person X and Person Y have to be either 5th or 6th in line. How many different orders are possible?

What is the probability of sitting next to a neighbor?

With probability 4 5 the man sits in one of the eight seats that have two neighbors, and in that case his wife’s probability of ending up next to him is 2 9. The overall probability that the end up sitting together is therefore 1 5 ⋅ 1 9 + 4 5 ⋅ 2 9 = 9 45 = 1 5. Added: And just for fun, here’s yet another.

How many different ways CAN YOU line up the women?

In how many different ways can you line up the women? As there are only three women, it is not difficult to ponder a moment and write down all the possibilities; referring to the ladies by the letters A, B, C, we discover the six arrangements ABC ACB BAC BCA CAB

What is the probability of seating 10 people in a room?

Since there are 10! ways to seat 10 people, the answer is 1/5. As a sanity check, here is another derivation. We concern ourselves only with the two people in the couple. There is a 8 / 10 chance that the first person will not be on either end, and given this a 2 / 9 chance that the second person will be next to them.

How many questions are there in the a quiz?

A quiz consists of four true-false questions and four multiple-choice questions with five choices each. How many different sets of answers are there? How many two-letter codes could be constructed from the first eleven letters of the Greek alphabet if no repetitions are not allowed?

How many people do I need to line up for tickets?

A group of 10 people need to line up for concert tickets. There are 6 spaces in the line. Person x and person y will be either 2nd or 3rd in the line. How many different orders are possible?

How many ways can five different door prizes be distributed?

How many ways can five different door prizes be distributed among five people? There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 · 4 · 3 · 2 · 1 = 120 ways. Now we will consider some slightly different examples. Example 27