<PRE><B>Question 321191</B>
a combination lock has digits from 0 to 39. three numbers are required to open.
how many combinations are possible if you cannot repeat any digits?
40*39*38 = 59280
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how many combination are possible if you cannot have the same digit twice in a row?
Call the number F, S and T for 1st, 2nd and 3rd
F is 1/40
S is any but F, = 1/39
T is any but S, = 1/39
40*39*39 = 60840

this is what i came up with..59,280 for first one and can&#39;t figure out second one:(</PRE>