. 3 Upper C 1
A. 3
2. The third element in row 6 of​ Pascal's triangle.
B. 6
3. 14 Upper C 1
C. 6 Upper C 2
4. The last term in the expansion of left parenthesis x plus 1 right parenthesis Superscript 7
D. font size decreased by 1 left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 7 2nd Row 1st Column 6 EndMatrix right parenthesis
 
5. font size decreased by 1 left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 7 2nd Row 1st Column 1 EndMatrix right parenthesis
 
E. 1
6. font size decreased by 1 left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 4 2nd Row 1st Column 2 EndMatrix right parenthesis
 
F. 14

It looks like you have a series of combinatorial problems related to Pascal's triangle and binomial expansions. Let’s break down each part:

1. **3 Choose 1 (3C1)**: This represents the number of ways to choose 1 element from a set of 3, calculated as:
   \[
   \binom{3}{1} = 3
   \]
   **Answer**: A. 3

2. **The third element in row 6 of Pascal's triangle**: The elements in row \( n \) are given by \( \binom{n}{k} \), where \( k = 0, 1, 2, \ldots, n \). The third element corresponds to \( k = 2 \):
   \[
   \binom{6}{2} = \frac{6!}{2!(6-2)!} = 15
   \]
   It seems there might be a mistake in your options if you're considering the third element as starting from \( k = 0 \). Assuming it's zero-based indexing, the answer is 15.

3. **14 Choose 1 (14C1)**: This represents choosing 1 element from a set of 14:
   \[
   \binom{14}{1} = 14
   \]
   **Answer**: F. 14

4. **The last term in the expansion of \((x + 1)^7\)**: The last term corresponds to \( k = 7 \) (the maximum power):
   \[
   \text{Last term} = \binom{7}{7} x^0 \cdot 1^7 = 1
   \]
   **Answer**: E. 1

5. **\(\binom{7}{1}\)**: This represents choosing 1 element from 7:
   \[
   \binom{7}{1} = 7
   \]

6. **\(\binom{4}{2}\)**: This represents choosing 2 elements from 4:
   \[
   \binom{4}{2} = \frac{4!}{2!2!} = 6
   \]

Now, let's summarize the answers based on the interpretations:

1. A. 3
2. B. 15 (assuming \( k=2 \))
3. F. 14
4. E. 1
5. 7 (not listed)
6. 6

If you have specific questions or need further clarification, let me know!

### Questions for Further Exploration:
1. How is the binomial coefficient calculated for larger numbers?
2. What are the applications of Pascal's triangle in combinatorics?
3. Can you derive the binomial expansion formula?
4. How does the symmetry in Pascal's triangle work?
5. What is the significance of the last term in a binomial expansion?

### Tip:
When calculating binomial coefficients, remember that \( \binom{n}{k} = \binom{n}{n-k} \). This can often simplify calculations!

3 Upper C 1; The third element in row 6 of Pascal's triangle; 14 Upper C 1; The last term in the expansion of (x + 1)^7; font size decreased by 1 ( Start 2 By 1 Matrix 1st Row 1st Column 7 2nd Row 1st Column 6 EndMatrix ) ; font size decreased by 1 ( Start 2 By 1 Matrix 1st Row 1st Column 7 2nd Row 1st Column 1 EndMatrix ) ; font size decreased by 1 ( Start 2 By 1 Matrix 1st Row 1st Column 4 2nd Row 1st Column 2 EndMatrix )

Explore combinatorial problems involving Pascal's Triangle and binomial expansions. Solve for 3C1, find the third element in row 6 of Pascal's triangle, and discover the last term in (x + 1)^7. Ideal for high school students.

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