Algebra II - Other Factorials

What is the value of ?

Explanation

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

Thus, .

Evaluate:

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

$1\frac{17}{24}$

$2 \frac{1}{12}$

$1\frac{3}{4}$

$1\frac{1}{33}$

None of the other choices gives the correct response.

Explanation

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

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

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

Explanation

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 .

Try without a calculator:

is equal to which expression?

$7! \cdot 8$

$7! \cdot 7$

None of these

Explanation

- 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.

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

Explanation

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:

Try without a calculator.

Give the value of that makes this three-part inequality true.

The inequality has no solution.

Explanation

Given a nonnegative integer , - or factorial - is defined to be the product of the integers from 1 to . After some exploration, multiplying 1 by 2, the result by 3, that result by 4, and so forth, we find that

$7! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5,040$

Multiplying by 8, we find that

$8! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8= 40,320$

Multiplying by 9, we find that

$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9= 362,880$

Therefore,

Since only integers may have factorial values, it follows that the only value of that makes the inequality true is .

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

$\textup{The answer is not given.}$

Explanation

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:

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

$\frac{3}{5}$

$\frac{3}{4}$

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{5}$

Explanation

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: $\frac{3}{5}$

Try without a calculator:

True or false: $\sqrt{9!}= 3!$

False

True

Explanation

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

$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 362,880$

$3! = 1\times 2 \times 3 = 6$

Since $6^{2} = 36$ ,

$(3!)^{2} = 36 e 362,880 = 9!$

and

$\sqrt{9!}= 3!$ is a false statement.

Try without a calculator:

Calculate:

Cannot be determined

Explanation

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

$3! = 1\times 2 \times 3 = 6$

Therefore,

Since

$6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$ ,

it follows that 720 is the correct response.

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