Question: Using operators plus, minus, multiplied by and divide by, and as many brackets as you want, can you do a formula which uses 3, 3, 5 and 7 to make 24? 

Each number must be used and can only be used once (so there will be two 3's).

So, for example, (3x7)+3 makes 24, but this isn't valid because the 5 wasn't used.

I don't know if this is possible, btw! It's in a game I'm playing.

Answer: 
We can use as many brackets as we want, and [Wikipedia states](https://en.wikipedia.org/wiki/Bracket_(mathematics)#Floor/ceiling_functions_and_fractional_part)

> 
> Square brackets, as in [π] = 3, are sometimes used to denote the floor function, which rounds a real number down to the next integer. 
> 
> 
> 

So I will use square brackets `[]` to denote the floor function.

> 
>  $$([5 \div 3]+7) \times 3 = (1+7) \times 3 = 8 \times 3 = 24$$
> 
> 
> 

If we massage the rules a little more, and can put numbers in different positions, we can use [Falling Factorial](https://en.wikipedia.org/wiki/Bracket_(mathematics)#Falling_and_rising_factorial) notation, where

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

> 
>  $$[(5)\_3 \div 7] \times 3 = \big[\frac{5!}{(5-3)!} \div 7 \big] \times 3 = [60 \div 7] \times 3 = 8 \times 3 = 24$$
> 
> 
> 

Unfortunately, there is no way to use [combinations](https://en.wikipedia.org/wiki/Binomial_coefficient) to get the given input to 24.
I also thought about : 

> 
>  $3 \* (7 - 5) ^ 3$
> 
> 
> 

Breaks the rule but hey