Factorials and lists/subsets counting

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SUMMARY

The discussion focuses on solving two mathematical problems involving factorials and permutations. The first problem calculates the value of ##\frac{120!}{118!}##, yielding a result of 14280 through the simplification of the factorial expression. The second problem involves counting the number of 9-digit numbers formed from the digits 1-9, with the condition that all odd digits precede even digits. The correct approach involves calculating the permutations of odd and even digits separately, resulting in a total of 2880 combinations using the formula ##5! \cdot 4!##.

PREREQUISITES
  • Understanding of factorial notation and operations (e.g., ##n!##)
  • Basic knowledge of permutations and combinations
  • Familiarity with odd and even number properties
  • Ability to perform arithmetic operations with factorials
NEXT STEPS
  • Study the properties and applications of factorials in combinatorics
  • Learn about permutations with restrictions, such as ordering conditions
  • Explore advanced counting techniques, including the use of binomial coefficients
  • Practice solving similar problems involving factorials and permutations
USEFUL FOR

Students studying combinatorics, educators teaching factorial concepts, and anyone interested in solving mathematical problems involving permutations and counting principles.

reenmachine
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1.1 Homework Statement

Using only pencil and paper , find the value of ##\frac{120!}{118!}##

2.1 Relevant equations

##\frac{120 \cdot 119 \cdot 118!}{118!} = 120 \cdot 119 = 14280##

1.2 Homework Statement

Compute how many 9-digit numbers can be made from the digits 1,2,3,4,5,6,7,8,9 if repetition is not allowed and all the odd digits occur first (on the left) followed by all the even digits (i.e. as in 1375980264 but not 0123456789).

(this question is from a book so I have no clue why they included 0 in their exemples , so I will just ignore it and take for granted that it's a mistake)

2.2 Relevant equations

My guess is this is like doing two different lists , one for the odd numbers and one for the even numbers.

The formula I intuitively came up with is ##5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 1## which in retrospective is simply ##5!4! = 2880##.

Any thoughts on those two problems and my attempt at solving them?

Thanks!
 
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All looks right.
 
thank you!
 

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