Factorials - Example and Practice Problems - Neurochispas (2024)

Solution

We can evaluate this by fully expanding the factorial:

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

$latex =40 320$

We can see that we got a large number. Factorials grow rapidly.

Solution

We have to expand the factorials to simplify:

$latex \frac{6!}{4!}=\frac{1 \times 2\times 3\times 4 \times 5\times 6}{1 \times 2\times 3\times 4}$

We see that the numbers from 1 to 4 are repeated both in the numerator and in the denomination, so we can eliminate them:

$latex\frac{1 \times 2\times 3\times 4 \times 5\times 6}{1 \times 2\times 3\times 4}= 5\times 6$

$latex =30$

Solution

Expanding these factorials, we have:

$latex \frac{5!}{2!3!}=\frac{5\times 4\times 3\times 2\times 1}{(2\times 1)(3\times 2\times 1)}$

We can simplify the numbers from 1 to 3. We can also simplify the 4 with the 2:

$latex\frac{5\times 4\times 3\times 2\times 1}{(2\times 1)(3\times 2\times 1)}=10$

Solution

We start by expanding the factorials:

$latex \frac{17!}{14!3!}=\frac{1\times 2\times 3\times 4…14\times 15 \times 16 \times 17}{(1\times 2\times 3\times 4 …14)(1\times 2 \times 3)}$

Now, we can simplify the numbers from 1 to 14 that are common to both 14! as for 17 !:

$latex\frac{1\times 2\times 3\times 4…14\times 15 \times 16 \times 17}{(1\times 2\times 3\times 4 …14)(1\times 2 \times 3)}=\frac{ 15 \times 16 \times 17}{1\times 2 \times 3}$

Now, we can simplify to 2 and 3 with 16 and 15:

$latex\frac{ 15 \times 16 \times 17}{1\times 2 \times 3}=5\times 8\times 17=680$

In this exercise, we use the three dots in the middle. This is useful to simplify the factorial expression a bit.

Solution

The factorial expression in the numerator is greater than the factorial expression in the denominator, so we expandn! partially until the expression $latex (n-2)!$ appears:

$latex \frac{n!}{(n-2)!}=\frac{n(n-1)(n-2)!}{(n-2)!}$

Now, we can cancel the common factors:

$latex\frac{n(n-1)(n-2)!}{(n-2)!}=n(n-1)$

We can apply the distributive property to simplify:

$latex n(n-1)={{n}^2}-n$

Solution

In this case, the denominator is clearly greater than the numerator since 3 is being added tonas opposed to the numerator which is only being added by 1.

We keep the numerator unchanged while expanding the denominator until the expression $latex (k + 1)!$ appears:

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

Now, we cancel the common factors:

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

We multiply the binomials in the denominator to finish:

$latex\frac{1}{(k+3)(k+2)}=\frac{1}{{{k}^2}+5k+6}$

Solution

Here, the numerator is greater since we are adding 2 and in the denominator we are subtracting 1. Therefore, we partially expand the numerator until the expression $latex (n-1)!$ appears:

$latex\frac{(n+2)!}{(n-1)!}=\frac{(n+2)(n+1)(n)(n-1)!}{(n-1)!}$

We simplify $latex (n-1)!$ both in the numerator and in the denominator:

$$\frac{(n+2)(n+1)(n)(n-1)!}{(n-1)!}=(n+2)(n+1)(n)$$

Now, we expand the expression to simplify:

$latex(n+2)(n+1)(n)={{n}^3}+3{{n}^2}+2n$

Find the result of 6!.

Choose an answer

120

540

720

840

Find the result of $latex \frac{6!}{3!}$.

Choose an answer

80

120

240

360

Simplify the expression $latex \frac{5!}{2!3!}$.

Choose an answer

10

12

15

20

Simplify the expression $latex \frac{x!}{(x-1)!}$.

Choose an answer

$latex (x+1)$

$latex x$

$latex (x-1)$

$latex (x-2)$