Let A denote the event that a 6-digit integer formed from 0,1,…,6 without repetition is divisible by 3. Find P(A).
Solutions · 1
Official solution
Total 6-digit numbers (no leading 0) from 7 digits =6⋅6!. Sum of all 7 digits =21. Removing a digit divisible by 3 keeps sum divisible by 3: remove 0, 3, or 6.
Case remove 3 (digits {0,1,2,4,5,6}): 5⋅5!; remove 6 ({0,1,2,3,4,5}): 5⋅5!; remove 0 ({1,2,3,4,5,6}): 6!.
\[P = \frac{6!+5\cdot5!+5\cdot5!}{6\cdot6!} = \frac{720+600+600}{4320}=\frac{1920}{4320}=\frac49\]