Math Problem Statement

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.

Solution

The question asks to evaluate two expressions involving factorials, given specific values of nn and rr:

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

n!(nr)!\frac{n!}{(n-r)!}

with n=11n = 11 and r=5r = 5.

This expression simplifies as follows:

11!(115)!=11!6!\frac{11!}{(11-5)!} = \frac{11!}{6!}

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

11×10×9×8×7×6!6!=11×10×9×8×7\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:

11×10=11011 \times 10 = 110 110×9=990110 \times 9 = 990 990×8=7920990 \times 8 = 7920 7920×7=55,4407920 \times 7 = 55,440

So, the value of 11!6!=55,440\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 n!r!(nr)!\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.

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Math Problem Analysis

Mathematical Concepts

Permutations
Factorials
Combinatorics

Formulas

n!/(n-r)!
n!/r!(n-r)!

Theorems

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Suitable Grade Level

High School