Factorial Number Exercises

factor numbers are positive integers that indicate the product between the number itself and all its predecessors.

For $\dpi{120} n\geq 2$ , We have to:

$\dpi{120} \boldsymbol{n! = n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdot ...\cdot 2\cdot 1}$

For $\dpi{120} n = 0$ and $\dpi{120} n =1$ , the factorial is defined as follows:

$\dpi{120} \boldsymbol{0! = 1}$
$\dpi{120} \boldsymbol{1!=1}$

To learn more about these numbers, see a list of factorial number exercises, all with resolution!

Index

Factorial Number Exercises
Resolution of question 1
Resolution of question 2
Resolution of question 3
Resolution of question 4
Resolution of question 5
Resolution of question 6
Resolution of question 7
Resolution of question 8

Factorial Number Exercises

Question 1. Calculate the factorial of:

a) 4
b) 5
c) 6
d) 7

Question 2. Determine the value of:

a) 5! + 3!
b) 6! – 4!
c) 8! – 7! + 1! – 0!

Question 3. Solve the operations:

a) 8!. 8!
b) 5! – 2!. 3!
c) 4!. (1 + 0)!

Question 4. Calculate the divisions between factorials:

The) $\dpi{120} \frac{10!}{9!}$

B) $\dpi{120} \frac{(10-4)!}{4!}$

ç) $\dpi{120} \frac{20!}{(19 + 1! - 0!)!}$

Question 5. Being $\dpi{120} a\in \mathbb{Z}$ , $\dpi{120} a> 0$ , express $\dpi{120} (a+5)!$ across $\dpi{120} a!$

Question 6. Simplify the following ratios:

The) $\dpi{120} \frac{(n+1)!}{n!}$

B) $\dpi{120} \frac{n!}{(n-1)!}$

ç) $\dpi{120} \frac{(n+3)!}{(n+3).(n+2).(n+1)}$

Question 7. Solve the equation:

$\dpi{120} 12x! + 5(x + 1)! = (x + 2)!$

Question 8. Simplify the quotient:

$\dpi{120} \frac{(x + 2)^3 \cdot x!}{(x+2)! + (x + 1)! + x!}$

Resolution of question 1

a) The factorial of 4 is given by:

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4! = 4. 3. 2. 1 = 24

b) The factorial of 5 is given by:

5! = 5. 4. 3. 2. 1

Like 4. 3. 2. 1 = 4!, we can rewrite 5! this way:

5! = 5. 4!

We've already seen that 4! = 24, so:

5! = 5. 24 = 120

c) The factorial of 6 is given by:

6! = 6. 5. 4. 3. 2. 1

Like 5. 4. 3. 2. 1 = 5!, we can rewrite 6! as follows:

6! = 6. 5! = 6. 120 = 720

d) The factorial of 7 is given by:

7! = 7. 6. 5. 4. 3. 2. 1

Like 6. 5. 4. 3. 2. 1 = 6!, we can rewrite 7! this way:

7! = 7. 6! = 7. 720 = 5040

Resolution of question 2

a) 5! + 3! = ?

When adding or subtracting factorial numbers, we must calculate each factorial before performing the operation.

Like 5! = 120 and 3! = 6, so we have to:

5! + 3! = 120 + 6 = 126

b) 6! – 4! = ?

Like 6! = 720 and 4! = 24, we have to:

6! – 4! = 720 – 24 = 696

c) 8! – 7! + 1! – 0! = ?

Like 8! = 40320, 7! = 5040, 1! = 1 and 0! = 1, we have to:

8! – 7! + 1! – 0! = 40320 – 5040 + 1 – 1 = 35280

Resolution of question 3

a) 8!. 8! = ?

In the multiplication of factorial numbers, we must calculate the factorials and then perform the multiplication between them.

Like 8! = 40320, so we have to:

8!. 8! = 40320. 40320 = 1625702400

b) 5! – 2!. 3! = ?

Like 5! = 120, 2! = 2 and 3! = 6, we have to:

5! – 2!. 3! = 120 – 2. 6 = 120 – 12 = 108

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c) 4!. (1 + 0)! = 4!. 1! = ?

Like 4! = 24 and 1! = 1, so we have to:

4!. 1! = 24. 1 = 24

Resolution of question 4

The) $\dpi{120} \frac{10!}{9!}$ = ?

In dividing factorial numbers, we must also calculate the factorials before solving the division.

Like 10! = 3628800 and 9! = 362880, so, $\dpi{120} \frac{10!}{9!} = \frac{3628800}{362880} = 10$ .

However, in division, we can simplify the factorials, canceling out equal terms in the numerator and denominator. This procedure facilitates many calculations. Look:

Like 10! = 10. 9. 8. 7. 6. 5. 4. 3. 2. 1 = 10. 9!, we have to:

$\dpi{120} \frac{10!}{9!} = \frac{10\cdot \cancel{9!}}{\cancel{9!}} = 10$

B) $\dpi{120} \frac{(10-4)!}{4!}$ = ?

$\dpi{120} \frac{(10-4)!}{4!} = \frac{6!}{4!} = \frac{6\cdot 5\cdot \cancel{4!}}{\cancel {4!}} = 30$

ç) $\dpi{120} \frac{20!}{(19 + 1! - 0!)!}$ = ?

$\dpi{120} \frac{20!}{(19 + 1! - 0!)!} = \frac{20!}{(19 + 1 - 1)!} = \frac{20!}{19!} = \frac{20\cdot \cancel{19!}}{\ cancel{19!}} = 20$

Resolution of question 5

Remembering that $\dpi{120} n! = n. (n - 1)!$ , we can rewrite $\dpi{120} (a+5)!$ this way:

$\dpi{120} (a+5)! = (a + 5). (a + 5 - 1)! = (a + 5). (a + 4)!$

Following this procedure, we have to:

$\dpi{120} (a+5)! = (a + 5). (a + 4). (a + 3). (a+ 2). (a + 1). The!$

Resolution of question 6

The) $\dpi{120} \frac{(n+1)!}{n!}$ = ?

We can rewrite the numerator as follows:

$\dpi{120} (n+1)! = (n+1).(n+1 - 1)! = (n+1).n!$

In this way, we were able to cancel the term $\dpi{120} n!$ , simplifying the quotient:

$\dpi{120} \frac{(n+1)!}{n!} = \frac{(n+1).\cancel{n!}}{\cancel{n!}} = n+1$

B) $\dpi{120} \frac{n!}{(n-1)!}$ = ?

We can rewrite the numerator as follows:

$\dpi{120} n! = n.(n-1)!$

Thus, we were able to cancel the term $\dpi{120} n!$ , simplifying the quotient:

$\dpi{120} \frac{n!}{(n-1)!} = \frac{n. \cancel{(n-1)!}}{\cancel{(n-1)!}} = n$

ç) $\dpi{120} \frac{(n+3)!}{(n+3).(n+2).(n+1)}$ = ?

We can rewrite the numerator as follows:

$\dpi{120} (n+3)! = (n+3).(n+2).(n+1). no!$

Thus, we can cancel some terms from the quotient:

$\dpi{120} \frac{(n+3)!}{(n+3).(n+2).(n+1)}= \frac{\cancel{(n+3).(n+) 2).(n+1)}.n!}{\cancel{(n+3).(n+2).(n+1)}} = n!$

Resolution of question 7

solve the equation $\dpi{120} 12x! + 5(x + 1)! = (x + 2)!$ means finding the values of $\dpi{120} x$ for which equality is true.

Let's start by decomposing terms with factorials, in an attempt to simplify the equation:

$\dpi{120} 12x! + 5(x + 1)! = (x + 2)!$

$\dpi{120} \Rightarrow 12x! + 5(x + 1).x! = (x + 2).(x+1).x!$

dividing both sides by $\dpi{120} x!$ , we managed to eliminate the factorial from the equation:

$\dpi{120} \frac{12\cancel{x!}}{\cancel{x!}} + \frac{5(x + 1).\cancel{x!}}{\cancel{x!}} = \frac{(x + 2).(x+1).\cancel{x!}}{\cancel{x!}}$

$\dpi{120} \Rightarrow 12 + 5(x + 1) = (x + 2).(x+1)$

By multiplying the terms in parentheses and arranging the equation, we have to:

$\dpi{120} 12 + 5x + 5 = x^2 + x + 2x + 2$

$\dpi{120} x^2 - 2x - 15 = 0$

It is a 2nd degree equation. From the Bhaskara formula, we determine the roots:

$\dpi{120} x = 5 \, \mathrm{or}\, x = -3$

By definition of factorial, $\dpi{120} x$ can't be negative, so, $\dpi{120} x = 5$ .

Resolution of question 8

$\dpi{120} \frac{(x + 2)^3 \cdot x!}{(x+2)! + (x + 1)! + x!}$

Like $\dpi{120} (x+2)! = (x+2).(x+1).x!$ and $\dpi{120} (x+1)! = (x+1).x!$ , we can rewrite the quotient as:

$\dpi{120} \frac{(x + 2)^3 \cdot x!}{(x+2).(x+1).x! + (x + 1).x! + x!}$

As the three portions of the denominator have the term $\dpi{120} x!$ , we can highlight it and cancel with $\dpi{120} x!$ that appears in the numerator.