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Question
A five-digit number is written down at random. The probability that the number is divisible by 5, and no two consecutive digits are identical, is
Options
\[\frac{1}{5}\]
\[\frac{1}{5} \left( \frac{9}{10} \right)^3\]
\[\left( \frac{3}{5} \right)^4\]
None of these
Solution
Let number be abcde
Case 1 : e = 0
a, b, c can be filled in 9 × 9 × 9 ways
c = 0 ⇒ 9 × 8 × 1 ways and d has 9 choices
c ≠ 0 ⇒ (9 × 9 × 9 – 9 × 8 × 1) = 657
in the case d has 8 choices ⇒ 657 × 8
Total case = 9 × 8 × 1 × 9 + 657 × 8 ⇒ 5904
Case 2 : e = 5
If c = 5,
if a ≠ 5 then a, b, c can be filled in 8 × 8 × 1 = 64 ways
if a = 5 then a, b, c can be filled in 1 × 9 × 1 = 9 ways
if c ≠ 5, then first 3 digits can be filled in 729 – 64 – 9 = 656 ways
here d has 8 choices
No. of member ending in 5 and no two consecutive digits being identical ⇒ (64 + 9) × 9 + 656 × 8
Hence, None of these
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