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Chapter 20: Problem 28

In how many ways can a committee consisting of 3 men and 4 women be formed from a group of 6 men and 7 women? (1) \({ }^{6} \mathrm{C}_{-}^{7} \mathrm{C}\) (2) \({ }^{6} \mathrm{C}^{7} \mathrm{C}\) (3) \({ }^{6} \mathrm{C}_{[}^{7} \mathrm{C}\) (4) ' \(\mathrm{C}^{6} \mathrm{C}\)

Short Answer

Expert verified
Answer: A committee consisting of 3 men and 4 women can be formed in 700 different ways.

Step by step solution

01

Combination Formula

To solve this problem we need to use the combination formula, which is \(\,_{n}C_{r}=\frac{n!}{r!(n-r)!}\) where \(_{n}C_{r}\) is the number of combinations of n elements taken r at a time, n is the total number of elements, r is the number of elements to choose, and ! denotes the factorial function.
02

Selecting 3 Men from 6 Men

Apply the combination formula to select 3 men from 6 men: \(\,_{6}C_{3}=\frac{6!}{3!(6-3)!}\)
03

Calculating the Number of Ways to Select 3 Men

Calculate the result: \(\,_{6}C_{3}=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1} = 20\) So, there are 20 ways to select 3 men from 6 men.
04

Selecting 4 Women from 7 Women

Apply the combination formula to select 4 women from 7 women: \(\,_{7}C_{4}=\frac{7!}{4!(7-4)!}\)
05

Calculating the Number of Ways to Select 4 Women

Calculate the result: \(\,_{7}C_{4}=\frac{7!}{4!3!}=\frac{7\times6\times5}{3\times2\times1} = 35\) So, there are 35 ways to select 4 women from 7 women.
06

Total Ways to Form the Committee

To find the total number of possible ways to form a committee consisting of 3 men and 4 women, multiply the number of ways of selecting men and women: Total ways = (Ways to select 3 men) 脳 (Ways to select 4 women) Total ways = 20 脳 35 = 700 Therefore, a committee consisting of 3 men and 4 women can be formed in 700 ways.

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Key Concepts

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

Combination Formula
The combination formula is a powerful tool in combinatorics. It helps determine the number of ways to select a group from a larger set, without caring about the order of selection. This is particularly useful when dealing with problems that involve group formation. The formula is given by: \[_{n}C_{r} = \frac{n!}{r!(n-r)!}\]In this formula,
  • \(n\) represents the total number of items
  • \(r\) is the number of items we want to choose
  • \(!\) is the factorial symbol, which multiplies a series of descending natural numbers
To solve any problem involving combinations, simply plug in the values of \(n\) and \(r\) into the formula. This will give you the number of ways to choose \(r\) items from \(n\).
Factorial Concept
Factorials are a fundamental concept in mathematics, often used within the combination formula. The notation \( n! \) denotes a factorial, which is the product of all positive integers up to \(n\). For example,
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
  • \(3! = 3 \times 2 \times 1 = 6\)
  • \(1! = 1\)
  • By definition, \(0! = 1\)
Factorials help calculate permutations and combinations. In the combination formula, they represent the total ways to arrange \(n\) items, the number of ways to arrange \(r\) items, and the arrangements of items not chosen. Always simplify your factorial calculations, as they can grow large quickly.
Committee Selection
Committee selection problems are a common application of combinatorics. They involve selecting a subgroup from a larger set, just like forming a committee from a group of people. This specific problem illustrates how to select members based on specific requirements, such as choosing 3 men from 6 and 4 women from 7.To solve such problems, follow these steps:
  • Use the combination formula to calculate the number of ways to choose the required number of members from each subset.
  • For example, choose 3 men from 6, which is calculated as \(\,_{6}C_{3}\), and 4 women from 7, which is \(\,_{7}C_{4}\).
  • Multiply the number of ways to get the total number of ways to form the committee.
The final calculation reveals how many different committee configurations meet the set criteria.
Mathematical Problem Solving
Mathematical problem solving with combinatorics often involves a step-by-step approach to break down complex problems into more manageable parts. Here's a simple guide to tackling such problems:
  • Clearly understand what is being asked. Identify the elements and the selections needed.
  • Recognize whether the problem involves combination (order doesn't matter) or permutation (order matters).
  • Write down the appropriate formula, like the combination formula in this case.
  • Substitute the given numbers into the formula to calculate.
  • For multiple requirements in a problem, calculate each separately and ensure to combine them correctly, often by multiplication.
Practice is key in mastering mathematical problem solving. Approach similar problems iteratively, using these strategies to build confidence and skill. This will help you solve them more efficiently over time.

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Most popular questions from this chapter

Using all the letters of the word 'QUESTION', how many different words can be formed? (1) \(8 !\) (2) \(7 !\) (3) \(7 \times 7 !\) (4) \(9 !\)

Using the digits \(0,1,2,5\) and 7 how many 4 -digit numbers that are divisible by 5 can be formed if repetition of the digits is not allowed? (1) 38 (2) 46 (3) 32 (4) 42

Tweleve teams are participating in a cricket tournament. If every team plays exactly one match with every other team, then the total number of matches played in the tournament is (1) 132 (2) 44 (3) 66 (4) 88

Anil has 8 friends. In how many ways can he invite one or more of his friends to a dinner? (1) 127 (2) 128 (3) 256 (4) 255

The number of diagonals of a regular polygon is 14 . Find the number of the sides of the polygon. (1) 7 (2) 8 (3) 6 (4) 9
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