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1 digit - 4 ways (any one can be chosen)

2 - digits - 4*3 ways (you picked the first one, so now you have 3 to choose from)

Waaait, isn't it 4C1

Answer is supposed to be 64.

For which one?

ehhh... im lost.. crap i suck at probability

Let me call amistre

yay amistre's here

So would one possibility be 1278?

One digit possibilities: 1, 2, 7 and 8. So there are 4 permutations for one digits...?

I understand now!

4 possibilities for 1 digits. 4x3 = 12 possibilities for 2 digits. And so on..

yes that's what i was trying to say adnan

lol, almost had that right

60+4 if i learn to count :)

yes lol

Any formula i could have used to save me time?

not really; since it is split up into 4 groups to count individually you still have a bit of work

knowing the cPr stuff helps tho

nPr ... cant type

I understood the formula before. But now, I just find it confusing.

I see. I work it out intuitively now - probably takes me a bit longer, but i get there eventually.