A qualifying string contains somewhere between m = 0 and n-3 “1”s. These m 1s may be placed in any of n positions. So total number of strings T becomes:
T = sum( {m=0 -> n-3} (n)! / [(n-m)! m!] )
We are summing over binary coefficients (n over m) usually called (n choose m)
I have verified this result using strings, and it also agrees with C and D’s solutions. While neither as direct nor as easy a route as theirs, it works (and may be readily generalized to any number of zeros)!
Edited on October 3, 2018, 6:03 pm
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