# Intro to associative property ofÂ multiplication

CCSS Math: 3.OA.B.5
Practice changing the grouping of factors in multiplication problems and see how it affects the product.

## Grouping numbers

The image shows $\tealD{3}$ rows with $\goldD{2}$ dots in each row. We can use the expression $\tealD{3} \times \goldD{2}$ to represent the array.
This image shows the same $\tealD{3} \times \goldD{2}$ array copied $\purpleD{4}$ times.
We use the expression $(\tealD{3} \times \goldD{2}) \times \purpleD{4}$ to represent the array.
The parentheses tell us which numbers to multiply first.
In the problem $(\tealD{3} \times \goldD{2}) \times \purpleD{4}$ we would multiply $\tealD{3} \times \goldD{2}$ first because those two numbers are inside the parentheses.
If we count the dots, we get a total of $24$.

### Changing the grouping

Will we get the same total if we change the parentheses so the numbers are grouped in a different way?
Let's regroup the numbers so the $\goldD{2}$ and the $\purpleD{4}$ are grouped together: $\tealD{3} \times (\goldD{2}\times \purpleD{4})$.
We can also draw an array to represent this expression. Let's start with $\goldD{2}$ rows with $\purpleD{4}$ dots in each row. This array shows $\goldD{2} \times\purpleD{4}$.
Now we need to copy the array $\tealD{3}$ times to represent the expression $\tealD{3} \times (\goldD{2} \times \purpleD{4})$.
If we count the dots, we still get a total of $24$.
Regrouping does not change the answer!
$(\tealD{3} \times \goldD{2}) \times \purpleD{4} = \tealD{3} \times (\goldD{2} \times \purpleD{4})$

## Associative property

The math rule that allows us to regroup numbers in a multiplication problem without changing the answer is the associative property.
Let's group the numbers in the following multiplication problem two different ways and show that we get the same product both ways.
$5 \times 4 \times 2$
Let's start by grouping the $\blueD{5}$ and the $\blueD{4}$ together. We can evaluate the expression step by step.
$\phantom{=}(\blueD{5 \times 4}) \times 2$
$= \blueD{20} \times 2$
$= 40$
Now let's group the $\purpleD{4}$ and the $\purpleD{2}$ together.
$\phantom{=}5 \times (\purpleD{4 \times 2})$
$=5 \times \purpleD{8}$
$=40$
We got the same product even though the numbers were grouped two different ways.
All three expressions are equal:
$\phantom{=}5 \times 4 \times 2$
$=(\blueD{5 \times 4}) \times 2$
$=5 \times (\purpleD{4 \times 2})$

### Let's try a few problems

Problem 1
Which expressions are equal to $6 \times 3 \times 4$?
Now let's try evaluating an expression two different ways.
Problem 2
Fill in the missing information to solve the expression $(\purpleD{3 \times 2}) \times 5.$
$(\purpleD{3 \times 2}) \times 5$$~=~$
$\times\, 5$
$\phantom{(\purpleD{3 \times 2}) \times 5}$$~=~$

Now solve the same expression that has been grouped in a different way.
Problem 3
Fill in the missing information to solve the expression $3 \times (\greenD{2 \times 5})$.
$3 \times (\greenD{2 \times 5})$$~=~ 3 \times$

$\phantom{3 \times (\greenD{2 \times 5})}$$~=~$

$(\purpleD{3 \times 2}) \times 5 = 30$ and
$3 \times (\greenD{2 \times 5}) = 30$
We got the same product even though we grouped the numbers two different ways.

### Equivalent expressions

We can use the associative property to find expressions that are equivalent.
Let's start with the expression $2 \times 2 \times 5$.
We can group this expression two ways that are both equivalent to $2 \times 2 \times 5$:
$(\blueD{2 \times 2}) \times 5$
$2 \times (\goldD{2 \times 5})$
By evaluating each expression step by step we can find other expressions that are also equivalent.
$(\blueD{2 \times 2}) \times 5 = \blueD{4} \times 5$
$2 \times (\goldD{2 \times 5}) = 2 \times \goldD{10}$
So our original expression, $2 \times 2 \times 5$, is also equivalent to $4 \times 5$ and $2 \times 10$.
Problem 4
Which expressions are equivalent to $8 \times 2 \times 4$?

## Why regroup?

Regrouping can make solving a multiplication problem easier.
Letâ€™s look at the expression, $4 \times 4 \times 5$.
We can group the expression two ways:
$(4 \times 4) \times 5$
$4 \times (4 \times 5)$
If we evaluate the first expression step by step we get: $(\blueD{4\times 4}) \times 5 = \blueD{16} \times 5$
If we evaluate the second expression step by step we get: $4 \times (\purpleD{4 \times 5}) = 4 \times \purpleD{20}$
It might be easier to find the product of $4 \times 20$ than $16 \times 5$.
Even though the numbers were grouped differently, both expressions have the same product.
$4 \times 20 = 80$
$16 \times 5 = 80$

### Let's try a problem

Problem 5
How can we group the expression $2 \times 3 \times 9$?