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Teammates. Four couples at a dinner party play a board game after the meal. They decide to play as teams of two and to select the teams randomly. All eight people write their names on slips of paper. The slips are thoroughly mixed, then drawn two at a time. How likely is it that every person will be teamed with someone other than the person he or she came to the party with?

Short Answer

Expert verified
The probability is approximately 8.57%.

Step by step solution

01

Understanding the Problem

We have 8 people at the party, divided into 4 couples. We want to find the probability that when these 8 people are randomly paired into 4 teams, no one ends up in a team with their original partner.
02

Calculate Total Number of Pairings

Calculate the total number of ways to pair 8 people into 4 teams of 2. This is given by the formula:\[\frac{8!}{(2!)^4 \times 4!}\]where \(8!\) is the total number of ways to arrange 8 people, \((2!)^4\) accounts for the ways to arrange the members within each team, and \(4!\) accounts for the arrangement of the 4 teams. Calculating gives \(\frac{40320}{384} = 105\).
03

Calculate Pairings with Restrictions

Use the principle of inclusion and exclusion to find the number of pairings where no one is paired with their partner. We define a derangement where no member of any pair is grouped with their original partner. Calculate the derangement for 4 couples (pairs):\[D_n = 4! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right)\]Calculate this to obtain \(9\).
04

Calculate Probability

The probability that no one is teamed with their partner is the number of valid pairings divided by the total number of pairings:\[P = \frac{9}{105} = \frac{3}{35} \approx 0.0857\]
05

Conclusion

Thus, the probability that each person is paired with someone other than their partner is approximately \( \frac{3}{35} \), or about 8.57%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arranging, and combining items. It's like solving puzzles where you determine different ways to arrange certain objects. In the context of our board game exercise, combinatorics helps us calculate the total number of possible pairings of people into teams.

When we need to pair 8 people into 4 teams, we use a formula that accounts for all arrangements and also for the indistinguishable nature of team members. The formula \[ \frac{8!}{(2!)^4 \times 4!} \]takes the total number of ways to arrange the 8 people \((8!)\) and adjusts it. The division by \((2!)^4\) handles the internal arrangement of each team, while \(4!\) addresses the distinct ordering of the teams themselves. This combination gives us \(105\) possible groupings.
  • The factorial \(8!\) is the total possible arrangement of 8 distinct individuals.
  • Each team member can be exchanged within the team, reducing arrangements by \((2!)\).
  • Teams themselves can be arranged, affecting total combinations, hence division by \(4!\).
Understanding these basic principles helps build a strong intuition about how we can handle and dissect large sets of data into understandable and countable parts.
Derangement
Derangement refers to a scenario where none of the objects appear in their original position when rearranged. Imagine mixing up a group of items such that all are misplaced. This concept is crucial in our problem, ensuring no individual is paired with their original companion at the party.

In our exercise, a derangement specifically applies to ensuring four couples are entirely mixed, meaning each person is paired with someone who is not their partner. The formula for a derangement is derived as follows:\[D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + \frac{(-1)^n}{n!}\right)\]which, when calculated for the 4 pairs, gives us \(9\) valid pairings that meet this misalignment condition.
  • \(n!\): Total arrangements of \(n\) items.
  • The series corrects for overly general counting by subtraction and addition.
  • Negative and positive terms adjust for double counting errors by permuting all but one pair at each step.
Derangements provide a neat way to ensure all the excitement of random pairing, fully avoiding expected or 'forbidden' matches.
Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a systematic method used to calculate the size of a union of overlapping sets. It subtracts overcounted elements while ensuring every unique element is counted correctly. In this problem, it helps us count pairings that abide by specific restrictions, such as avoiding a person being placed with their original date.

This principle works by initially allowing for all possible counts, then subtracting overlapping cases that don't meet criteria, before adding back in any cases subtracted too many times. Applying it in our context:
  • Consider overlapping constraints of pairing problems.
  • Use the formula to enumerate combinations that meet conditions.
  • Prune invalid overlaps to restore correct count balance.
In essence, the formula used for derangements in our exercise relies heavily on the Inclusion-Exclusion Principle to ensure only the desired outcomes are left in the count. By meticulously handling overlaps, we can accurately segment and sanction mix-ups, leading to the calculated success probability of \(\frac{9}{105}\) where none end up with their partner.

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Most popular questions from this chapter

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Beat the lottery. Many states run lotteries to raise money. A Web site advertises that it knows "how to increase YOUR chances of Winning the Lottery." They offer several systems and criticize others as foolish. One system is called Lucky Numbers. People who play the Lucky Numbers system just pick a "lucky" number to play, but maybe some numbers are luckier than others. Let's use a simulation to see how well this system works. To make the situation manageable, simulate a simple lottery in which a single digit from 0 to 9 is selected as the winning number. Pick a single value to bet, such as 1 , and keep playing it over and over. You'll want to run at least 100 trials. (If you can program the simulations on a computer, run several hundred. Or generalize the questions to a lottery that chooses two- or three-digit numbers-for which you'll need thousands of trials.) a) What proportion of the time do you expect to win? b) Would you expect better results if you picked a "luckier" number, such as \(7 ?\) (Try it if you don't know.) Explain.

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