Design a unique and personalized textbook problem where at least one item must be chosen from a group of items where order does not matter and there are no identical elements (Lesson 5.3). Solve the problem using two different methods that were taught in the course notes. Explain why both methods give the same answer. write it about a stamp collection, use permutations to solve once, and bionomeol theorom to solve again then explain why its the same answer

### Textbook Problem: Stamp Collection Selection

**Problem Statement:**

Emily has a collection of 7 unique stamps from different countries. She wants to choose a subset of these stamps for a display. 

1. How many ways can she choose at least one stamp from her collection?

### Solution:

#### Method 1: Using Permutations

First, let's calculate the number of ways to choose any subset of stamps. Since the order of selection does not matter and there are no identical elements, we can use combinations.

For \( n = 7 \) stamps, the total number of subsets (including the empty set) is \( 2^n \). However, we need to exclude the empty set since Emily wants to choose at least one stamp.

\[ \text{Total subsets} = 2^n = 2^7 = 128 \]

\[ \text{Total non-empty subsets} = 2^7 - 1 = 128 - 1 = 127 \]

Thus, there are 127 ways to choose at least one stamp from her collection.

#### Method 2: Using the Binomial Theorem

The Binomial Theorem states that for any non-negative integer \( n \):

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

Setting \( x = 1 \) and \( y = 1 \), we get:

\[ (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} \]

For \( n = 7 \):

\[ 2^7 = \sum_{k=0}^{7} \binom{7}{k} \]

This represents the sum of all possible ways to choose 0 to 7 stamps. To find the number of ways to choose at least one stamp, we subtract the case where no stamps are chosen (i.e., \( \binom{7}{0} = 1 \)):

\[ 2^7 - \binom{7}{0} = 128 - 1 = 127 \]

Thus, using the binomial theorem, we again find there are 127 ways to choose at least one stamp from her collection.

#### Explanation: Why Both Methods Give the Same Answer

Both methods ultimately count the number of non-empty subsets of the set of 7 unique stamps. The permutations method leverages the principle of counting all possible subsets and subtracting the empty set, while the binomial theorem method sums the binomial coefficients representing all possible selections of stamps and again subtracts the empty set case.

The reason they provide the same answer is that both approaches fundamentally rely on the same combinatorial principles of counting subsets. In permutations, we consider the power set (all subsets), and in the binomial theorem, we sum over all possible choices for each subset size. Both methods are just different ways to approach the same combinatorial problem.

### Questions for Further Exploration

1. How many ways can Emily choose exactly 3 stamps from her collection of 7?
2. If Emily had 10 unique stamps, how many ways can she choose at least one stamp?
3. How many ways can she choose at least 2 stamps from her collection of 7?
4. If she selects 4 stamps, how many possible combinations can be made?
5. How does the number of ways change if she decides to include an additional stamp, making her collection 8 stamps in total?

### Tip

When solving combinatorial problems, always check if the order of selection matters. If it does, use permutations; if it doesn’t, use combinations. This helps in choosing the right method to approach the problem.

Design a unique and personalized textbook problem where at least one item must be chosen from a group of items where order does not matter and there are no identical elements (Lesson 5.3). Solve the problem using two different methods that were taught in the course notes. Explain why both methods give the same answer. Write it about a stamp collection, use permutations to solve once, and binomial theorem to solve again, then explain why it's the same answer.

Explore how to solve a stamp collection selection problem using both permutations and the binomial theorem. This problem highlights key concepts in combinatorics, with a detailed explanation of why both methods yield the same answer. Suitable for Grades 9-12 students studying permutations and combinations.

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