unfavourable outcomes : [which is 1XXXX and X1XXX], i’m getting 12 each[4!/2! each. to find the favourable outcomes, isn’t this what greg does?].
total : 60 as determined, so wouldn’t favourable outcomes be = (60 - 24)/60?
I think i’m making some mistake in the unfavourable outcomes. could someone please help me on that?
That is right so far, but you’re missing these cases for your unfavorable outcomes:
XX1XX
XXX1X
ah okay. so it’s actually simpler to go by greg’s method.
in any case, could you help me with what the results of these unfavourable outcomes would be? is it 2![XX] x 1 X 1[because both 7s after the 1] for the XX1XX
and
XXX1X → 4!/2!?
Both 7s don’t have to be after 1. Notice how something like 27137 also works as a complement cuz this is an arrangement where 2 7’s are not to the left of 1. Try accounting for this in the 2 cases you haven’t worked out.
Also there’s a super easy way to solve it without any computation.
