Problem 32
Question
Find the number of ways a committee of five can be formed from a group of five boys and four girls, if each committee must contain: At least two boys.
Step-by-Step Solution
Verified Answer
The total number of ways to form a committee of five containing at least two boys from a group of five boys and four girls is 121.
1Step 1: Identify the possible combinations of boys and girls
At least two boys are needed in each committee. Therefore, the possible combinations are:
1. Two boys, three girls (2B3G),
2. Three boys, two girls (3B2G),
3. Four boys, one girl (4B1G),
4. Five boys (5B).
2Step 2: Calculate the number of ways for each combination
We will use the combinations formula \(C(n,k)=\frac{n!}{k!(n-k)!}\) to calculate the number of ways to form each combination, where n is the total number of people in the group (boys or girls), and k is the number of people to choose from the group:
1. 2B3G: C(5, 2) * C(4, 3) = \(\frac{5!}{2!(5-2)!}\) * \(\frac{4!}{3!(4-3)!}\) = 10 * 4 = 40
2. 3B2G: C(5, 3) * C(4, 2) = \(\frac{5!}{3!(5-3)!}\) * \(\frac{4!}{2!(4-2)!}\) = 10 * 6 = 60
3. 4B1G: C(5, 4) * C(4, 1) = \(\frac{5!}{4!(5-4)!}\) * \(\frac{4!}{1!(4-1)!}\) = 5 * 4 = 20
4. 5B: C(5, 5) = \(\frac{5!}{5!(5-5)!}\) = 1
3Step 3: Sum the number of ways for all combinations
To find the total number of ways to form a committee, we will now sum up the number of ways previously calculated for each combination:
40 (2B3G) + 60 (3B2G) + 20 (4B1G) + 1 (5B) = 121
The total number of ways to form a committee of five containing at least two boys is 121.
Key Concepts
Permutation and CombinationFactorial NotationBinomial Coefficients
Permutation and Combination
Understanding permutation and combination is crucial for solving problems related to counting and probability in discrete mathematics. Permutations refer to the arrangements of items in a specific order, which makes order important. Combinations, on the other hand, are selections of items regardless of the order, which means that the order does not matter.
For instance, if we have a set of three letters {A, B, C}, the permutations of these three letters taken two at a time are AB, BA, AC, CA, BC, CB, constituting six permutations. However, when we consider combinations, the pairs AB and BA are not distinct as the order does not matter; hence we only have three combinations: AB, AC, and BC.
In the given exercise, we looked at forming committees (which is a selection problem and thus related to combinations), where the order of selection does not matter. We calculated the number of ways to select groups of boys and girls without regard to the order in which they are chosen. This is key to understanding why we used the combination formula and not the permutation formula.
For instance, if we have a set of three letters {A, B, C}, the permutations of these three letters taken two at a time are AB, BA, AC, CA, BC, CB, constituting six permutations. However, when we consider combinations, the pairs AB and BA are not distinct as the order does not matter; hence we only have three combinations: AB, AC, and BC.
In the given exercise, we looked at forming committees (which is a selection problem and thus related to combinations), where the order of selection does not matter. We calculated the number of ways to select groups of boys and girls without regard to the order in which they are chosen. This is key to understanding why we used the combination formula and not the permutation formula.
Factorial Notation
Factorial notation is a mathematical concept that is widely used in permutations and combinations. The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n. Essentially, it's a shorthand way to express multiplications that have a specific pattern.
For example, the factorial of 5 (written as 5!) is calculated as:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorial notation becomes valuable when dealing with large numbers or when calculating permutations and combinations since it simplifies expressions and calculations. For instance, in our exercise, factorial notation allowed us to quickly find out the number of combinations without having to manually enumerate every possibility.
For example, the factorial of 5 (written as 5!) is calculated as:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorial notation becomes valuable when dealing with large numbers or when calculating permutations and combinations since it simplifies expressions and calculations. For instance, in our exercise, factorial notation allowed us to quickly find out the number of combinations without having to manually enumerate every possibility.
Binomial Coefficients
Binomial coefficients are a central element of combinatorics and can be visually represented in Pascal's Triangle. They are often denoted as \( C(n, k) \) or \( \binom{n}{k} \), which represent the number of ways to choose k items from a set of n distinct items without regard to the order of selection.
These coefficients are named for their role in the binomial theorem and are also the coefficients in the expansion of \( (x + y)^n \) when expanded using binomial theory. We often calculate these using the formula involving factorials:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
The binomial coefficients were used in the provided solution to calculate the different ways to form committees. It's important to understand that these coefficients are symmetric, meaning \( C(n, k) \) is the same as \( C(n, n-k) \), and this property is sometimes useful in simplifying calculations or in probability theory.
These coefficients are named for their role in the binomial theorem and are also the coefficients in the expansion of \( (x + y)^n \) when expanded using binomial theory. We often calculate these using the formula involving factorials:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
The binomial coefficients were used in the provided solution to calculate the different ways to form committees. It's important to understand that these coefficients are symmetric, meaning \( C(n, k) \) is the same as \( C(n, n-k) \), and this property is sometimes useful in simplifying calculations or in probability theory.
Other exercises in this chapter
Problem 32
Using the binomial theorem, prove each. \(\sum_{r=0}^{n}\left(\begin{array}{c}n \\\ r\end{array}\right)\left(\begin{array}{c}n \\\ n-r\end{array}\right)\left(\b
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Find the number of solutions to each equation. $$x_{1}+x_{2}+x_{3}+x_{4}=11, x_{1}, x_{2} \geq 2,2 \leq x_{3} \leq 4, x_{4} \geq 3$$
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A botanist would like to plant three coleus, four zinnias, and five dahlias in a row in her front garden. How many ways can she plant them if: The family of zin
View solution Problem 32
Find the number of solutions to each equation. $$x_{1}+x_{2}+x_{3}+x_{4}=11, x_{1}, x_{2} \geq 2,2 \leq x_{3} \leq 4, x_{4} \geq 3$$
View solution