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Evaluate each factorial expression. \(\frac{4 ! 11 !}{7 !}\)

Short Answer

Expert verified
190080

Step by step solution

01

Write out the factorials

First, write out the factorials explicitly. Recall that the factorial of a number is the product of all positive integers less than or equal to that number.For example, \[ 4! = 4 \times 3 \times 2 \times 1 \ 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]
02

Simplifying the expression

Substitute the factorials into the expression \[ \frac{4! \times 11!}{7!} \]This yields the following:\[ \frac{(4 \times 3 \times 2 \times 1) \times (11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \]
03

Cancel common terms

Observe that the terms \(7!\) in the numerator and denominator will cancel each other out.So, the expression reduces to:\[ \frac{4! \times 11!}{7!} = (4 \times 3 \times 2 \times 1) \times (11 \times 10 \times 9 \times 8) \]
04

Compute the remaining factorials and product

Evaluate the factorial for 4! and then multiply the remaining terms for 11! from 11 to 8.First, compute \(4!\):\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]Then, compute the product of the remaining terms of 11! from 11 to 8:\[ 11 \times 10 \times 9 \times 8 = 7920 \]Multiply the results together:\[ 24 \times 7920 = 190080 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factorials
Factorials are a fundamental concept in mathematics, especially in combinatorics and algebra.
A factorial is the product of all positive integers up to a certain number.
We denote factorials with an exclamation mark (!). For example, 5! (read as 'five factorial') is: \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorials grow very fast. For large numbers, they can become extremely big.
Understanding factorials and how to compute them is crucial in solving problems related to permutations, combinations, and various algebraic expressions.
simplifying expressions
Simplifying expressions is about reducing them to their simplest form.
This often involves canceling common terms, combining like terms, and performing basic arithmetic.
In our problem, we start with the expression \[\frac{4! \times 11!}{7!}\].
We write out the factorials: \[ 4! = 4 \times 3 \times 2 \times 1 \11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \].
By cancelling the common terms, we simplify the expression to \[(4 \times 3 \times 2 \times 1) \times (11 \times 10 \times 9 \times 8)\].
Simplification helps us see the problem more clearly and reduces the amount of computation needed.
multiplication of integers
Multiplication of integers is a basic arithmetic operation that combines two numbers to produce a product.
It's fundamental in all areas of math, including when dealing with factorials.
When multiplying several integers, it's helpful to break them down into smaller steps to avoid mistakes.
For example, we had to compute \[4! = 4 \times 3 \times 2 \times 1 = 24\] and then \[11 \times 10 \times 9 \times 8 = 7920 \].
Finally, by multiplying 24 and 7920 together, we get \[24 \times 7920 = 190080 \].
Knowing how to handle multiplication precisely is essential for solving factorial expressions and other algebraic problems efficiently.

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