Probability of Occupying Adjacent Seats in a Random Arrangement

In summary, the conversation discusses the probability of k people occupying k adjacent seats in a row with n seats. The probability is found by dividing the number of sets of k adjacent seats by the total number of seat combinations. The conversation also mentions the equation (n choose k) = n!/[(n-k)!k!] and how to simplify the expression (n-k+1)(n-k)!k!. The solution is found by recognizing that (n-k+1) is one more than (n-k), leading to the conclusion that (n-k+1)(n-k)! is equal to (n-k+1)!.
  • #1
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Homework Statement



If k people are seated in a random manner in a row containing n seats (n>k), what is the probability that the people will occupy k adjacent seats in the row?

I realize that there are n choose k sets of k seats to be occupied, and that there are n-k+1 sets of k adjacent seats. So the probability that I'm looking for is:

(n-k+1)/(n choose k)

What I don't understand is how the above probability simplifies to:

[(n-k+1)!k!]/n!

Can someone please explain? Thanks.

Homework Equations



(nk) = n choose k = n!/[(n-k)!k!]

The Attempt at a Solution



(n-k+1)/(n choose k) = [(n-k+1)(n-k)!k!]/n!

Not sure what to do from here.
 
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  • #2
Look at (n-k+1)(n-k)!, can you simplify this product any?
 
Last edited:
  • #3
Other than finding a quotient, I've never had to manipulate/simplify factorials. Thinking about this again, since the number (n-k+1) is 1 greater than the number (n-k), (n-k)! times (n-k+1) must equal (n-k+1)!.
 

Related to Probability of Occupying Adjacent Seats in a Random Arrangement

1. What is a basic combinations problem?

A basic combinations problem is a mathematical problem that involves finding the number of ways a certain set of items can be arranged or selected without repetition. It is also known as a "counting" problem.

2. How do you solve a basic combinations problem?

To solve a basic combinations problem, you can use the formula nCr = n! / (r!*(n-r)!), where n represents the total number of items and r represents the number of items being selected. Alternatively, you can use a combination table or tree diagram to list out all the possible combinations.

3. What is the difference between a combination and a permutation?

A combination is a selection of items from a set without regard to the order in which they are selected, while a permutation is a rearrangement of items in a set where the order matters. For example, selecting 3 out of 5 objects without regard to order would be a combination, while arranging 3 objects in a specific order would be a permutation.

4. Can a basic combinations problem have repetition?

No, a basic combinations problem does not involve repetition. This means that once an item is selected, it cannot be selected again in the same combination. If repetition is allowed, it becomes a permutations problem instead.

5. In what real-life situations can basic combinations problems be used?

Basic combinations problems can be used in various real-life situations, such as determining the number of possible outcomes in a lottery, calculating the number of possible seating arrangements at a dinner party, or finding the number of possible combinations on a lock with multiple digits.

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