The 4-aces is a 3u probability question and aren't asked in the mathematics exam..

You can always work towards the binomial probability taught in 3u from mathematics level anyway:

A = ace, O = other

you can either get AAAAO, AAAOA, AAOAA, AOAAA and OAAAA...

so with 5 combinations:

P(4 aces) = {(4/52) * (3/51) * (2/50) * (1/49) * (48/48)} + {(4/52) * (3/51) * (2/50) * (48/49) * (1/48)} + ...

at this point, better 2u students should see the pattern:

the probability of getting 4 aces in the 1st order is the same as the 2nd..

and as there are 5 orders it can be achieved, just take the first order and multiply by 5.

either that or they can just keep going for all the orders