Suppose the group of twelve consists of five men and seven women. a) How many five-person teams can be chosen that consist of three men and two women? b) How many five-person teams contain at least one man? c) How many five-person teams contain at most one man?
Suppose the group of twelve consists of five men and seven women. a) How many five-person teams can be chosen that consist of three men and two women? b) How many five-person teams contain at least one man? c) How many five-person teams contain at most one man?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Permutations and Combinations
If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.
Counting Theory
The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.
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![**Problem Statement:**
Suppose the group of twelve consists of five men and seven women.
a) How many five-person teams can be chosen that consist of three men and two women?
b) How many five-person teams contain at least one man?
c) How many five-person teams contain at most one man?
**Detailed Instructions:**
To solve these problems, we will employ combinatorics, specifically the combination formula, often noted as C(n, k), which calculates the number of ways to choose k items from a total of n items without regard to order. The formula is expressed as:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
Where \( n! \) denotes the factorial of n, which is the product of all positive integers up to n.
**Solution Steps:**
**a) Teams with three men and two women:**
1. Calculate the number of ways to choose 3 men from 5.
\[
C(5, 3) = \frac{5!}{3! \times (5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
2. Calculate the number of ways to choose 2 women from 7.
\[
C(7, 2) = \frac{7!}{2! \times (7-2)!} = \frac{7 \times 6}{2 \times 1} = 21
\]
3. The total number of teams is the product of these two values.
\[
C(5, 3) \times C(7, 2) = 10 \times 21 = 210
\]
**b) Teams with at least one man:**
1. Calculate the total number of five-person teams possible from 12 people.
\[
C(12, 5) = \frac{12!}{5! \times (12-5)!} = 792
\]
2. Calculate the number of all-woman teams (no men, five women).
\[
C(7, 5) = \frac{7!}{5! \times (7-5)!} = \frac{7 \times 6}{2 \times 1} = 21
\]
3. Subtract](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4640cd63-ee34-4c34-940b-cd98e9fa837c%2F9b6b8a90-2751-4808-82d6-ebe38329aafa%2Fkt4k9eb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose the group of twelve consists of five men and seven women.
a) How many five-person teams can be chosen that consist of three men and two women?
b) How many five-person teams contain at least one man?
c) How many five-person teams contain at most one man?
**Detailed Instructions:**
To solve these problems, we will employ combinatorics, specifically the combination formula, often noted as C(n, k), which calculates the number of ways to choose k items from a total of n items without regard to order. The formula is expressed as:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
Where \( n! \) denotes the factorial of n, which is the product of all positive integers up to n.
**Solution Steps:**
**a) Teams with three men and two women:**
1. Calculate the number of ways to choose 3 men from 5.
\[
C(5, 3) = \frac{5!}{3! \times (5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
2. Calculate the number of ways to choose 2 women from 7.
\[
C(7, 2) = \frac{7!}{2! \times (7-2)!} = \frac{7 \times 6}{2 \times 1} = 21
\]
3. The total number of teams is the product of these two values.
\[
C(5, 3) \times C(7, 2) = 10 \times 21 = 210
\]
**b) Teams with at least one man:**
1. Calculate the total number of five-person teams possible from 12 people.
\[
C(12, 5) = \frac{12!}{5! \times (12-5)!} = 792
\]
2. Calculate the number of all-woman teams (no men, five women).
\[
C(7, 5) = \frac{7!}{5! \times (7-5)!} = \frac{7 \times 6}{2 \times 1} = 21
\]
3. Subtract
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