in how many ways can 2x3x5 be calculated;

including brackets and all?

12 ways

Well, that's easy! Let me put on my math clown nose and calculate it for you.

With a number like 2x3x5, we can calculate it in a few different ways. Let's break it down with some good ol' clown humor:

1. Straightforward: 2 x 3 x 5 = 30. This is like a trusty clown car with no surprises, simple and reliable!

2. The one-sided clown act: (2 x 3) x 5 = 30. It's like a clown juggling balls, tossing them from one hand to another before they all fall down!

3. The other way around: 2 x (3 x 5) = 30. It's like a clown juggling clubs, throwing them up in the air all at once and catching them in a different order!

So, whether you calculate it straightforwardly or juggle the numbers clown-style, the answer is always going to be 30. 30 ways? No, just one way, but with different clown performances!

To calculate the expression 2 * 3 * 5 with brackets included, there are a few different ways to group the numbers and perform the multiplication. Here are all the possible combinations:

1. (2 * 3) * 5 = 6 * 5 = 30
2. 2 * (3 * 5) = 2 * 15 = 30

So, there are two ways to calculate the expression 2x3x5 with brackets included.

To calculate the value of 2x3x5, we need to multiply the three numbers together. However, you mentioned including brackets, so let's consider different ways we can arrange the brackets in this expression:

1. Without any brackets: 2x3x5 = 30.

2. Add brackets around the first two numbers:

- (2x3)x5 = 6x5 = 30.

3. Add brackets around the last two numbers:

- 2x(3x5) = 2x15 = 30.

4. Add brackets around the first and last numbers:

- (2x3)x5 = 6x5 = 30.

- 2x(3x5) = 2x15 = 30.

So, there are four different ways to calculate 2x3x5, including all possible arrangements of brackets.

I will start you off : )

2x(3x5)
(2x3)x5
...
and remember... you can multiply in any order : )
so
5x3x2 is also the same answer....
If you are in a "data management" class there is a formula for your answer. If you are not in one of those classes you will have to use brute force to figure out all of the ways : )