Math Problem Statement

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solve it more clearly

Solution

Let's solve the first problem clearly.

Problem 1:

A committee of 6 is to be chosen from 10 men and 7 women such that it contains at least 3 men and 2 women. However, two particular women refuse to serve on the same committee. How many different ways can this be done?

Step-by-step solution:

Step 1: Understanding the conditions

  • The committee consists of 6 people.
  • The committee must contain at least 3 men and 2 women.
  • There are 10 men and 7 women available.
  • Two particular women cannot serve on the same committee.

Step 2: Identify possible gender combinations for the committee

Since the committee must contain at least 3 men and 2 women, the valid combinations of men and women are:

  • 3 men and 3 women.
  • 4 men and 2 women.
  • 5 men and 1 woman.

Step 3: Calculate the number of ways to choose men and women for each combination

  1. For 3 men and 3 women:

    • Number of ways to choose 3 men from 10 men = (103)\binom{10}{3}.
    • Number of ways to choose 3 women from 7 women = (73)\binom{7}{3}.

    However, two women refuse to serve together. Let's denote them as W1W_1 and W2W_2. We need to subtract the cases where both W1W_1 and W2W_2 are chosen together.

    • Number of ways to choose 3 women without restriction = (73)\binom{7}{3}.
    • Number of ways to choose 3 women where both W1W_1 and W2W_2 are included: We must select 1 more woman from the remaining 5 (since W1W_1 and W2W_2 are already chosen), which is (51)\binom{5}{1}.

    Hence, the total number of valid ways to choose 3 women is: (73)(51)\binom{7}{3} - \binom{5}{1}

  2. For 4 men and 2 women:

    • Number of ways to choose 4 men from 10 men = (104)\binom{10}{4}.
    • Number of ways to choose 2 women from 7 women = (72)\binom{7}{2}.

    We need to exclude the cases where W1W_1 and W2W_2 are both chosen.

    • Number of ways to choose 2 women without restriction = (72)\binom{7}{2}.
    • Number of ways to choose 2 women where W1W_1 and W2W_2 are both chosen = 1 (since they are the only 2 women).

    Hence, the total number of valid ways to choose 2 women is: (72)1\binom{7}{2} - 1

  3. For 5 men and 1 woman:

    • Number of ways to choose 5 men from 10 men = (105)\binom{10}{5}.
    • Number of ways to choose 1 woman from 7 women = (71)\binom{7}{1}.

    In this case, there are no restrictions on the woman chosen because only 1 woman is selected, so we don't have to worry about W1W_1 and W2W_2 both being chosen.

Step 4: Calculate the total number of ways for each scenario

  1. 3 men and 3 women: (103)×((73)(51))\binom{10}{3} \times \left( \binom{7}{3} - \binom{5}{1} \right)

  2. 4 men and 2 women: (104)×((72)1)\binom{10}{4} \times \left( \binom{7}{2} - 1 \right)

  3. 5 men and 1 woman: (105)×(71)\binom{10}{5} \times \binom{7}{1}

Step 5: Add the results from all the scenarios

The total number of ways to form the committee is the sum of the valid choices from all three cases.

If you'd like, I can calculate the numerical values for you or walk you through each combination calculation. Would you like more details?

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Probability
Set Theory

Formulas

nCr = n! / (r!(n-r)!)
Conditional combinations

Theorems

Combination theorem

Suitable Grade Level

Undergraduate level, Discrete Mathematics