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Chapter 3: Problem 3

How many seven-element subsets are there in a set of nine elements?

Short Answer

Expert verified
There are 36 seven-element subsets in a set of nine elements.

Step by step solution

01

Understand the Problem

We need to find the number of seven-element subsets that can be formed from a set containing nine elements. This is a typical combinatorics problem where we use combinations.
02

Recall the Combination Formula

The formula to find the number of combinations (subsets of a certain size) is given by \( \binom{n}{k} = \frac{n!}{k! \cdot (n-k)!} \), where \(n\) is the total number of elements and \(k\) is the number of elements to choose.
03

Identify \(n\) and \(k\)

Here, the total number of elements \(n = 9\) and we need to select \(k = 7\) elements.
04

Plug in Values into the Formula

Using the formula for combinations, substitute \(n = 9\) and \(k = 7\):\[ \binom{9}{7} = \frac{9!}{7! \cdot (9-7)!} = \frac{9 \times 8}{2!} \]
05

Perform the Calculation

First, calculate the factorial in the formula.- Compute the numerator: \(9 \times 8 = 72\)- Compute the denominator: \(2! = 2\).Thus, \( \binom{9}{7} = \frac{72}{2} = 36\).
06

Validate the Calculation

To ensure the calculation is correct, notice that choosing 7 elements out of 9 is equivalent to leaving out 2 elements out of 9. Thus, \( \binom{9}{7} = \binom{9}{2} \), which simplifies similarly to give 36.

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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 key tool in combinatorics and helps us determine how many ways we can select a subset of items from a larger set, without considering the order of selection. This formula is represented mathematically as:
  • \( \binom{n}{k} \)
This expression reads as "n choose k" and calculates the number of combinations possible. It is defined as:\[\binom{n}{k} = \frac{n!}{k! \cdot (n-k)!}\]In this formula:
  • \(n\) is the total number of elements in the set.
  • \(k\) is the number of elements we want to choose from the set.
  • \(!\) denotes a factorial, which we will explore further in the next section.
This formula is extremely useful when handling problems where the order of selection is irrelevant, as in the case of subsets.
Factorial
The factorial, denoted by an exclamation mark \(!\), is a product of all positive integers up to a given number. It is a fundamental part of the combination formula and is used to calculate permutations and combinations.For any positive integer \(n\), the factorial \(n!\) is computed as:
  • \(n! = n \times (n-1) \times (n-2) \times \cdots \times 1\)
For example:
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
  • \(3! = 3 \times 2 \times 1 = 6\)
Factorials grow very quickly as \(n\) becomes larger and play a crucial role in counting the total number of arrangements and selections.
Combinations
Combinations refer to the selection of items from a set where order does not matter. They differ from permutations, where order is important. Combinations are a fundamental concept in probability and statistics, and they are vital for calculating possibilities in various scenarios.The number of combinations can be calculated using the combination formula, where we disregard the order, focusing instead solely on choosing a specific number of elements. For example:
  • Choosing 7 elements from 9: \( \binom{9}{7} = \frac{9!}{7! \cdot 2!} \)
Calculating above yields:
  • \(\binom{9}{7} = \frac{72}{2} = 36\)
Combinations help us efficiently solve problems involving selection, especially when order is not a factor, such as picking teams or forming groups.
Subset Calculation
Subset calculation involves finding the number of subsets of a particular size that can be formed from a larger set. In the given problem, we determine how many seven-element subsets can be created from a set of nine elements.Using combinations is critical in subset calculation as they help us find the number of ways to select a particular subset. This is calculated using:
  • The combination formula \( \binom{n}{k} \).
Here:
  • \(n = 9\), the total set of elements.
  • \(k = 7\), the size of the subset we want to determine.
After performing the calculation:
  • \(\binom{9}{7} = 36\).
This means there are 36 different ways to form a seven-element subset from a nine-element set, demonstrating the power of combinatorial mathematics in solving real-world problems.

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Most popular questions from this chapter

Each of the four engines on an airplane functions correctly on a given flight with probability \(.99,\) and the engines function independently of each other. Assume that the plane can make a safe landing if at least two of its engines are functioning correctly. What is the probability that the engines will allow for a safe landing?

A symphony orchestra has in its repertoire 30 Haydn symphonies, 15 modern works, and 9 Beethoven symphonies. Its program always consists of a Haydn symphony followed by a modern work, and then a Beethoven symphony. (a) How many different programs can it play? (b) How many different programs are there if the three pieces can be played in any order? (c) How many different three-piece programs are there if more than one piece from the same category can be played and they can be played in any order?

A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?

Charles claims that he can distinguish between beer and ale 75 percent of the time. Ruth bets that he cannot and, in fact, just guesses. To settle this, a bet is made: Charles is to be given ten small glasses, each having been filled with beer or ale, chosen by tossing a fair coin. He wins the bet if he gets seven or more correct. Find the probability that Charles wins if he has the ability that he claims. Find the probability that Ruth wins if Charles is guessing.

How many ways can six indistinguishable letters be put in three mail boxes? Hint: One representation of this is given by a sequence |LL|L|LLL| where the |'s represent the partitions for the boxes and the L's the letters. Any possible way can be so described. Note that we need two bars at the ends and the remaining two bars and the six L's can be put in any order.
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