Give a combinational proof of the identity C(n + 10, 2) = C(n, 2) + C(10, 2) + 10n using the 6 of the statements below. Put the statements in correct order.



The first way to answer this question is C(n + 10, 2).



This is because you could pick two positions for the 1s from the first n positions or from the last 10 positions, or put 1 in the first n positions and the other in the last 10 positions.



This is because there are n + 10 positions, which you must choose two to put a 1.



A second answer to the question is C(n, 2) + C(10, 2) 10n.



This is because there are n+10 letters all together and you have two choices for each.



Consider the question: How many n-letter words contain 10 different letters?

To provide a combinatorial proof for the identity \( C(n + 10, 2) = C(n, 2) + C(10, 2) + 10n \), we can structure the reasoning using the provided statements. The key idea is to count the number of ways to choose 2 positions from \( n + 10 \) in two different ways, and show that both methods lead to the same result. Here is the correct order of the statements that lead to a proof:

1. **The first way to answer this question is \( C(n + 10, 2) \).**
   - This establishes the left-hand side of the identity. You are choosing 2 positions from \( n + 10 \) total positions.

2. **This is because there are \( n + 10 \) positions, which you must choose two to put a 1.**
   - This explains why the expression \( C(n + 10, 2) \) is used, by counting all possible ways to select 2 positions from \( n + 10 \) items.

3. **A second answer to the question is \( C(n, 2) + C(10, 2) + 10n \).**
   - Now we consider an alternate way of counting the number of ways to choose 2 positions, breaking it into cases. This is the right-hand side of the identity.

4. **This is because you could pick two positions for the 1s from the first \( n \) positions or from the last 10 positions, or put 1 in the first \( n \) positions and the other in the last 10 positions.**
   - This explains the decomposition: 
     - \( C(n, 2) \) counts the ways to choose both positions from the first \( n \),
     - \( C(10, 2) \) counts the ways to choose both positions from the last 10,
     - \( 10n \) counts the ways to choose one position from the first \( n \) and the other from the last 10.

5. **This is because there are \( n + 10 \) letters all together and you have two choices for each.**
   - This justifies why these different cases are mutually exclusive and collectively exhaustive.

6. **Consider the question: How many \( n \)-letter words contain 10 different letters?**
   - This is not directly relevant for this problem, so we discard this statement.

The correct order of statements is:

1. The first way to answer this question is \( C(n + 10, 2) \).
2. This is because there are \( n + 10 \) positions, which you must choose two to put a 1.
3. A second answer to the question is \( C(n, 2) + C(10, 2) + 10n \).
4. This is because you could pick two positions for the 1s from the first \( n \) positions or from the last 10 positions, or put 1 in the first \( n \) positions and the other in the last 10 positions.
5. This is because there are \( n + 10 \) letters all together and you have two choices for each.

Would you like further details on any specific part of this proof or explanation?

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Here are 5 follow-up questions related to this topic:

1. What is the general formula for \( C(n, 2) \) and how is it derived?
2. How does the combination formula \( C(n, 2) \) change if we add constraints on selecting the positions?
3. Can this combinatorial proof approach be applied to other identities involving combinations?
4. What happens to the identity if we replace the number 10 with a different constant?
5. How does this identity relate to Pascal’s identity in combinatorics?

**Tip:** Always ensure that different cases in a combinatorial proof are mutually exclusive and collectively exhaustive to avoid over-counting or missing cases.

Give a combinational proof of the identity C(n + 10, 2) = C(n, 2) + C(10, 2) + 10n using the 6 of the statements below. Put the statements in correct order.

Explore a combinatorial proof of the identity C(n + 10, 2) = C(n, 2) + C(10, 2) + 10n using key concepts in combinatorics and binomial coefficients. This detailed solution is suitable for high school students in Grades 10-12.

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