[ \binom{7}{2} \times 2! = \frac{7!}{2!(7-2)!} \times 2 = \frac{7 \times 6}{2 \times 1} \times 2 = 21 \times 2 = 42 ]

Your equation is correct. Here's a breakdown of the calculation for clarity:

1. The binomial coefficient \(\binom{7}{2}\) is defined as:
   \[
   \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!}
   \]
2. Simplifying \(\frac{7!}{2!5!}\):
   \[
   \frac{7!}{2!5!} = \frac{7 \times 6 \times 5!}{2! \times 5!} = \frac{7 \times 6}{2!}
   \]
3. Now, calculate \(2!\):
   \[
   2! = 2 \times 1 = 2
   \]
4. The binomial coefficient becomes:
   \[
   \frac{7 \times 6}{2} = \frac{42}{2} = 21
   \]
5. Finally, you multiply by \(2!\) (which is \(2\)):
   \[
   21 \times 2 = 42
   \]

Thus, the final result is \(42\).

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Would you like more details or have any questions?

Here are some related questions for further practice:
1. How would you compute \(\binom{8}{3} \times 3!\)?
2. What is the value of \(\binom{6}{2} \times 2!\)?
3. How do you calculate \(\binom{n}{r} \times r!\) for any general \(n\) and \(r\)?
4. What is the factorial of 7 and how do you simplify \( \frac{7!}{2!(7-2)!} \)?
5. How does the binomial coefficient change if you swap \(n\) and \(r\)?

**Tip**: The binomial coefficient formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) counts the number of ways to choose \(r\) elements from a set of \(n\) elements without regard to order.

\binom{7}{2} \times 2! = \frac{7!}{2!(7-2)!} \times 2 = \frac{7 \times 6}{2 \times 1} \times 2 = 21 \times 2 = 42

Learn how to calculate the binomial coefficient \binom{7}{2} and multiply it by 2 factorial. This detailed step-by-step solution covers key concepts in combinatorics and factorials, suitable for students in Grades 9-12.

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