Part 1: Scheduling the Performers

If one performer insists on being the last stand-up comic, the remaining 87 performers can be arranged in any order before this final performer. The number of ways to arrange $n$ items is given by $n!$ (n factorial).

$\text{Total ways} = 87!$

So, there are $87!$ ways to schedule the appearances of the remaining performers while keeping the last performer fixed.

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Part 2: Evaluate $\frac{20!}{13!}$

The factorial expression $\frac{20!}{13!}$ simplifies by canceling the terms of $13!$ in both numerator and denominator:

$\frac{20!}{13!} = 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14$

Now calculate step-by-step:

$20 \times 19 = 380$ $380 \times 18 = 6840$ $6840 \times 17 = 116280$ $116280 \times 16 = 1860480$ $1860480 \times 15 = 27907200$ $27907200 \times 14 = 390700800$

So,

$\frac{20!}{13!} = 390,700,800$

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Final Answers:

1. Number of ways to schedule the appearances: $87!$
2. Value of $\frac{20!}{13!}$: $390,700,800$

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Do you have any questions or would like further details? Here are some related questions for exploration:

1. What is the value of $87!$ as an approximate result?
2. How does the presence of one fixed position reduce the total permutations of 88 performers?
3. Can you generalize the formula for such problems with one or more fixed performers?
4. What is the mathematical significance of factorial growth?
5. How can factorials be applied in combinatorics to count combinations or subsets?

Tip: Use factorials and their properties to simplify large expressions efficiently by canceling terms.