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The question asks to evaluate two expressions involving factorials, given specific values of \( n \) and \( r \):

For part (a), we are asked to evaluate:

\[
\frac{n!}{(n-r)!}
\]

with \( n = 11 \) and \( r = 5 \).

This expression simplifies as follows:

\[
\frac{11!}{(11-5)!} = \frac{11!}{6!}
\]

We can calculate this by canceling out the common factorial terms in the numerator and denominator:

\[
\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6!} = 11 \times 10 \times 9 \times 8 \times 7
\]

Now we multiply:

\[
11 \times 10 = 110
\]
\[
110 \times 9 = 990
\]
\[
990 \times 8 = 7920
\]
\[
7920 \times 7 = 55,440
\]

So, the value of \( \frac{11!}{6!} = 55,440 \).

If you would like help with part (b), feel free to upload the image or describe the equation!

Would you like more details or have any questions? Here are some related questions you can explore:
1. How do factorials behave for very large numbers?
2. What is the difference between permutations and combinations?
3. How can factorials be used in probability theory?
4. How can you simplify other factorial-based expressions like \( \frac{n!}{r!(n-r)!} \)?
5. Why do factorials grow so quickly?

**Tip:** Cancel out common factorial terms when simplifying expressions with factorials to reduce large numbers quickly.

For the given values of n and r, evaluate (a) n!/(n-r)! and (b) n!/r!(n-r)! for n=11 and r=5.

Explore how to calculate factorial-based expressions for permutations and combinations when n=11 and r=5. This guide simplifies n!/(n-r)! and provides a step-by-step approach to understand factorial operations in combinatorics. Suitable for high school students learning advanced math topics.

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