Q53 / 110 · VITEEE 2024 · Mathematics
Two red counters, three green counters and four blue counters are placed in a row in random order. The probability that no two blue counters are adjacent is
- A$\frac{7}{99}$
- B$\frac{7}{198}$
- Correct answerC$\frac{5}{42}$
- DNone of these
Solution
Total counters $=9(2 R+3 G+4 B)$
Number of ways of placing 9 counters $=9$ ! and Number of ways of placing there 9 counters in such a way that no 2 blue counters are adjacent $=5!\times{ }^6 C_4 \times 4!$
$\begin{aligned}
\text { Required probability } & =\frac{5!\times{ }^6 C_4 \times 4!}{9!} \\
& =\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 9 \times 8 \times 7 \times 6}=\frac{5}{42}
\end{aligned}$
Hence, the correct answer is option: 'C'.