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.

Let's solve this problem step by step.

### Problem Restatement:
We have six married couples and four additional men, making a total of \( 12 + 4 = 16 \) 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:

\[
\text{Total ways} = \binom{16}{4}
\]

### 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: \( \binom{6}{1} = 6 \) ways.
- Choose 2 more people from the remaining 14 (excluding the couple selected): \( \binom{14}{2} \) ways.

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

\[
\text{Ways with 1 couple} = 6 \times \binom{14}{2}
\]

#### Case 2: Two couples
- Choose 2 couples from the 6 married couples: \( \binom{6}{2} = 15 \) 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:

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

### Step 3: Subtracting Unwanted Cases
The number of valid selections is:

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

### Step 4: Calculation

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

\[
\binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820
\]

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

\[
\binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91
\]

3. Calculate \( 6 \times \binom{14}{2} \):

\[
6 \times 91 = 546
\]

4. Add the cases for two couples:

\[
546 + 15 = 561
\]

5. Subtract this from the total:

\[
1820 - 561 = 1259
\]

### Final Answer:
The number of ways to select 4 people such that no couple is selected together is \( \boxed{1259} \).

Would you like any further details or have any questions?

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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 \( r \) people from \( n \) 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.

Explore how to count the number of ways to select 4 people from 6 married couples and 4 additional men, ensuring no couples are together. Includes step-by-step solution using combinatorial principles and the combination formula. Suitable for advanced high school students and beyond.

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