Math Problem Statement

Six married couple and 4 men are sitting in a room. Find the number of ways in which 4 people can be selected so that no couples are in the committe together.

Solution

Let's solve this problem step by step.

Problem Restatement:

We have six married couples and four additional men, making a total of 12+4=1612 + 4 = 16 people. We need to select 4 people such that no couple is selected together.

Step 1: Counting Total Possible Selections

The total number of ways to select 4 people from 16 (without any restrictions) is given by the combination formula:

Total ways=(164)\text{Total ways} = \binom{16}{4}

Step 2: Counting Selections that Include Couples

Next, we must subtract the cases where at least one couple is selected together.

Case 1: One couple and two other people

  • Choose 1 couple from the 6 married couples: (61)=6\binom{6}{1} = 6 ways.
  • Choose 2 more people from the remaining 14 (excluding the couple selected): (142)\binom{14}{2} ways.

So, the number of ways to select 4 people where exactly one couple is selected is:

Ways with 1 couple=6×(142)\text{Ways with 1 couple} = 6 \times \binom{14}{2}

Case 2: Two couples

  • Choose 2 couples from the 6 married couples: (62)=15\binom{6}{2} = 15 ways.
  • No additional people need to be selected because 4 people have already been chosen.

So, the number of ways to select 4 people with exactly two couples is:

Ways with 2 couples=15\text{Ways with 2 couples} = 15

Step 3: Subtracting Unwanted Cases

The number of valid selections is:

Valid selections=(164)[6×(142)+15]\text{Valid selections} = \binom{16}{4} - \left[ 6 \times \binom{14}{2} + 15 \right]

Step 4: Calculation

  1. Calculate (164)\binom{16}{4}:

(164)=16×15×14×134×3×2×1=1820\binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820

  1. Calculate (142)\binom{14}{2}:

(142)=14×132×1=91\binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91

  1. Calculate 6×(142)6 \times \binom{14}{2}:

6×91=5466 \times 91 = 546

  1. Add the cases for two couples:

546+15=561546 + 15 = 561

  1. Subtract this from the total:

1820561=12591820 - 561 = 1259

Final Answer:

The number of ways to select 4 people such that no couple is selected together is 1259\boxed{1259}.

Would you like any further details or have any questions?


Here are five questions that relate to this problem:

  1. How would the solution change if there were 5 couples and 6 additional people?
  2. What if we had to select 5 people instead of 4, still ensuring no couples are selected together?
  3. How does the problem change if the additional men are replaced with additional women?
  4. What is the general formula for selecting rr people from nn such that no couple is together?
  5. How would the approach change if we were allowed to select exactly one couple but not two?

Tip: When solving combinatorial problems, always consider breaking down the problem into smaller cases to avoid overcounting or undercounting.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Combinations
Counting principles

Formulas

Combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Theorems

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Suitable Grade Level

Advanced High School