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

   

Answer 1

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

⇒ 5905

​Total cases ⇒ 5904 + 5905 ⇒ 11809

\[\text{ Required Probability } = \frac{11809}{90000}\]

Hence, None of these