Confused on dice problem

[spice rock]

" A balanced die is tossed six times, and the number on the upper most face is recorded each
time. What is the probability that the numbers recorded are 1,2,3,4,5,and6 in any order?"

[spice rock]

I feel as i am doing something wrong because that would be a massive ass number

[limpid bison]

how many do we have? 6^6
That's the total, that's what should be in the denominator

[limpid bison]

It would indeed

Think of how many ways there are to get 1,2,3,4,5,6 in any order

[spice rock]

Its only that? Oh.. then how come sometimes we do like combinations/permutations in the bottom?

[limpid bison]

Not sure what you're referring to; it all depends on the specific problem

[spice rock]

Oh okay

So reading

[limpid bison]

I'm sure if you practice you'll get better at identifying what kind of thing should go where

[spice rock]

Theres 1 in 6 ways to get 1, 1/6 ways to get 2 etc?

[limpid bison]

Not quite

You might be thinking about it the wrong way

Say I tossed six dies and got 1,2,3,4,5,6 but not necessarily in that order. In how many ways can I line them up?

[spice rock]

Uh many wayd

[limpid bison]

By lining them up, I'm ordering them

So there's a first one, a second one, etc.

How many can be first?

Once you've answered that, answer how many can be second given that one is already first

You should be able to continue the reasoning and find the correct solution

[spice rock]

6 can be first

OHH so this isnt a combination one its a probabkity one

6•5•4•3•2•1

[limpid bison]

I mean it is a permutation thing

If you know about nPr, this is just 6P6

[spice rock]

Oh. I think i need to review some things then

OHH cause order matters

[limpid bison]

nPr isn't very useful anyway, you can pretty much always think in terms of n! and nCr, which is also known as the binomial coefficient

And in this case it's simply 6!

[spice rock]

Ah okay

I took any order has unordered shit

So i need to practicw! Thanks

.solved