Assume now that there are 12 people initially on the yacht with 6 women and 6 men, and that exploration teams consist of six people. How many exploration teams have more women than men?
Assume now that there are 12 people initially on the yacht with 6 women and 6 men, and that exploration teams consist of six people. How many exploration teams have more women than men?
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Assume now that there are 12 people initially on the yacht with 6 women and 6 men, and that exploration teams consist of six people. How many exploration teams have more women than men?
![**Example 2.3.5: The Exploration Parties**
Suppose there are 11 people aboard a yacht, 5 women and 6 men, and 4 of them take the dinghy to explore an island. How many exploration parties have at least two women?
A quick and dirty solution is this: choose 2 women from the 5 and then choose 2 others from the 9 remaining people. (This will certainly produce an exploration party with 2 or more women.) Applying the product rule, we know that this can be done in
\[
\binom{5}{2} \times \binom{9}{2} = 10 \times 36 = 360 \text{ ways.}
\]
// But is this correct?
Does each “way” produce a unique exploration party?
That number cannot be right because the total number of possible exploration parties is
\[
\binom{11}{4} = \frac{11!}{4! \times 7!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} = 11 \times 10 \times 3 = 330.
\]
For \( w = 0 \) to 4, the number of parties that contain **exactly** \( w \) women can be counted by determining the number of ways \( w \) can be chosen from the 5 women, and then choosing the rest of the party from the men; that is, \( (4 - w) \) men are chosen from the 6 men.
\[
\begin{array}{ccc}
w & 4 - w & \binom{5}{w} \times \binom{6}{4-w} \\
0 & 4 & 1 \times 15 = 15 \\
1 & 3 & 5 \times 20 = 100 \\
2 & 2 & 10 \times 15 = 150 \quad // \times 1 = 150 \\
3 & 1 & 10 \times 6 = 60 \quad // \times 3 = 180 \\
4 & 0 & 5 \times 1 = \frac{5}{\quad} \quad // \times 6 = 30 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4792360-22b4-40c0-a48a-02ce03c9f6f7%2F4d33bed4-ccd6-4ebc-9932-7bc8bd2fdb0a%2F035fn2_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 2.3.5: The Exploration Parties**
Suppose there are 11 people aboard a yacht, 5 women and 6 men, and 4 of them take the dinghy to explore an island. How many exploration parties have at least two women?
A quick and dirty solution is this: choose 2 women from the 5 and then choose 2 others from the 9 remaining people. (This will certainly produce an exploration party with 2 or more women.) Applying the product rule, we know that this can be done in
\[
\binom{5}{2} \times \binom{9}{2} = 10 \times 36 = 360 \text{ ways.}
\]
// But is this correct?
Does each “way” produce a unique exploration party?
That number cannot be right because the total number of possible exploration parties is
\[
\binom{11}{4} = \frac{11!}{4! \times 7!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} = 11 \times 10 \times 3 = 330.
\]
For \( w = 0 \) to 4, the number of parties that contain **exactly** \( w \) women can be counted by determining the number of ways \( w \) can be chosen from the 5 women, and then choosing the rest of the party from the men; that is, \( (4 - w) \) men are chosen from the 6 men.
\[
\begin{array}{ccc}
w & 4 - w & \binom{5}{w} \times \binom{6}{4-w} \\
0 & 4 & 1 \times 15 = 15 \\
1 & 3 & 5 \times 20 = 100 \\
2 & 2 & 10 \times 15 = 150 \quad // \times 1 = 150 \\
3 & 1 & 10 \times 6 = 60 \quad // \times 3 = 180 \\
4 & 0 & 5 \times 1 = \frac{5}{\quad} \quad // \times 6 = 30 \
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