Other Factorials - Algebra 2

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Question

What is the value of ?

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Answer

! is the symbol for factorial, which means the product of the whole numbers less than the given number.

Thus, .

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Question

What is $\frac{10!}{7!}$ ?

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Answer

Remember that factorial is defined as:

$n! = n \cdot (n-1)\cdot (n-2) \cdots \cdot (3)\cdot (2)\cdot (1)$

So using this definition,

And .

So $\frac{10!}{7!}$ = $\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

The 7, 6, 5, 4, 3, 2, and 1 all cancel from both the numerator and denominator. So we're left with just on top, which has a value of .

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Question

How many -permutations (without repetition) are there when taking numbers from the set of numbers $\begin{Bmatrix} 16, 11, 9, 7, 14, 12, 18, 23, 1 \end{Bmatrix}$ ?

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Answer

The elements of the set don't matter. Only the size of the set matters when determining permutations.

Our set contains 9 integers, so for the first number in our permutation, we have 9 choices.

After picking that number, because we're not allowed repetition, our second number is from 8 choices.

Our final number is from 7 choices.

Multiplying gives us .

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Question

What is the value of $\frac{10!}{7!}$ ?

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Answer

Since the factorial has the property

$\frac{n!}{(n+1)!}$

we can write as:

$10 \cdot 9 \cdot 8 \cdot (7!)$ .

Thus, our expression can be written as

$$\frac{10\cdot9\cdot8\cdot7!}{7!}$=10\cdot9\cdot8=720$

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Question

Solve the Quadratic Equation.

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Answer

First, move all terms to the left by subtracting the quantity on the right.

$= $4x^{2}$ - 12x + 5 = 0$

From here, factor the quadractic into two binomials and set each equal to zero.

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Question

What is the value of ?

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Answer

A factorial represents the product of all natural numbers less than a given number. Thus, which gives us .

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Question

What is ?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach .

So, is $3\cdot 2\cdot 1$ .

Multiplying that out, we get .

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Question

What is ?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach .

So, is $5\cdot 4\cdot 3\cdot 2\cdot 1$ .

Multiplying that out, we get .

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Question

What is $3! \cdot 4!$ ?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach .

So, is $3\cdot 2\cdot 1$ . Multiplying that out, we get .

We also have . That is $4\cdot 3\cdot 2\cdot 1$ or .

Since we are multiplying the factorial, we multiply and to get .

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Question

What is $\frac{6!}{4!}$ ?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach .

So, is $6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ . We also have . That is $4\cdot 3\cdot 2\cdot1$ .

Since we are dividing the factorial, we can cancel out some terms.

Both the numerator and denominator have $4\cdot 3\cdot 2\cdot 1$ , and we can cancel those out. We are left with $6\cdot 5$ or

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Question

What is $\frac{7!}{30}$ ?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach .

So, is $7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ . We are dividing by With careful inspection, can be broken down to $6\cdot5$ .

If we cancel that out with the ones in the numerator, we have $7\cdot 4\cdot 3\cdot 2\cdot 1$ or .

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Question

What is $\frac{x!}{(x+1)!}$ ?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach . Since we are dealing with variables, let's analyze each the numerator and the denominator. is definitely greater than in this situation because factorials are always positive numbers. If we took the difference between and we would get . This means that if we were to expand both the numerator and denominator, we cancel everything out except the extra term in the denominator which is .

$\frac{x!}{(x+1)!)}$=\frac{x\cdot 1}{(x+1)(x)(1)}$=\frac{1}{x+1}$

So final answer is $\frac{1}{x+1}$ .

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Question

What can be expressed as?

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach . Since we are dealing with variables, let's analyze them. is definitely greater than in this situation because factorials are always positive numbers. If we took the difference between and we would get . This means after , the next biggest value is . Since we have and the next value multiplied is , we can conclude that will work since it incorporates values multiplied and also .

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Question

Simplify.

$\frac{(x+2)!}{(x-2)!}$

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Answer

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach . Since we are dealing with variables, let's analyze each the numerator and the denominator. is definitely greater than in this situation because factorials are always positive numbers. If we took the difference between and we would get . This means that if we were to expand both the numerator and denominator, we cancel everything out except the extra terms in the numerator which are . So final answer is .

If you need convincing, let . So we have $\frac{(6+2)!}{(6-2)!}$=\frac{8!}{4!}$=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1}$ .

You can cancel out the $4\cdot 3\cdot 2\cdot1$ from top and bottom to get $8\cdot 7\cdot 6\cdot 5$ . We need to express them into expressions.

Since was , to get , you need to add to or

To get , you add to or

To get you subtract from or .

You still get the same answer of .

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Question

Simplify: $\frac{2! + 0!}{3!-0!}$

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Answer

Write out the terms of each factorial in expanded form. The only exception is the zero factorial, which is equal to one.

$\frac{2! + 0!}{3!-0!}$ =\frac{2\times 1+1}{3\times 2\times 1-1}$

Simplify the terms. The values in the numerator and denominator cannot cancel out!

$\frac{2\times 1+1}{3\times 2\times 1-1}$ = $\frac{2+1}{6-1}$ = $\frac{3}{5}$

The correct answer is: $\frac{3}{5}$

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Question

What is the value of ?

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Answer

Step 1: We must define factorial. Factorial is defined as , where

Step 2: Evaluate

$8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=40320$

$40320\equiv 8!$

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Question

Compute: $\frac{(3!)!+3(3!)}{3!}$

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Answer

Simplify the factorials in the numerator and denominator.

$\frac{(3!)!+3(3!)}{3!}$=\frac{(3 \times 2 \times 1)!+3(3 \times 2 \times 1)}{3 \times 2 \times 1}$

Simplify the terms on the top and bottom.

$\frac{(3 \times 2 \times 1)!+3(3 \times 2 \times 1)}{3 \times 2 \times 1}$ =\frac{(6)!+3(6)}{6}$

$\frac{(6)!+3(6)}{6}$ = $\frac{(6\times 5\times4\times3\times2\times1)+3(6)}{6}$

$$\frac{720+18}{6}$ = $\frac{738}{6}$ = 123$

The answer is:

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Question

Multiply: $[(2!)! \cdot $(1-1)!]^2$$

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Answer

Simplify all the terms in the parentheses first.

$(2!) = 2\times 1 =2$

This indicates that:

$(2!)! \cdot (1-1)! = 2\cdot 1 =2$

$[(2!)! \cdot $(1-1)!]^2$ = $[2]^2$ = 4$

The answer is:

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Question

Try without a calculator:

is equal to which expression?

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Answer

- or factorial - is defined to be the product of the integers from 1 to . Therefore,

$7 ! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$

and

$8 ! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$

$= (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7) \cdot 8$

$= 7! \cdot 8$ ,

the correct response.

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Question

Evaluate:

$\sum_${k=1}^{4}$ $\frac{1}{k!}$$

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Answer

$\sum_${k=1}^{4}$ $\frac{1}{k!}$$ is equal to the sum of the expressions formed by substituting 1, 2, 3 and 4, in turn, for in the expression $\frac{1}{k!}$ , as follows:

- or factorial - is defined to be the product of the integers from 1 to . Therefore, each term can be calculated by multiplying the integers from 1 to , then taking the reciprocal of the result.

:

$$\frac{1}{k!}$ = $\frac{1}{1!}$ = $\frac{1}{1}$ = 1$

:

$\frac{1}{k!}$ = $\frac{1}{2!}$ = $\frac{1}{1 \cdot 2}$ = $\frac{1}{2}$

:

$\frac{1}{k!}$ = $\frac{1}{3!}$ = $\frac{1}{1 \cdot 2 \cdot 3}$ = $\frac{1}{6}$

:

$\frac{1}{k!}$ = $\frac{1}{4!}$ = $\frac{1}{1 \cdot 2 \cdot 3 \cdot 4}$ = $\frac{1}{24}$

Add the terms:

$\sum_${k=1}^{4}$ $\frac{1}{k!}$ = 1 + $\frac{1}{2}$ + $\frac{1}{6}$ + $\frac{1}{24}$ = 1$\frac{17}{24}$$

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