Question

Numbers 1,2,3,4,5,6,7,8 are arranged in a random order. The probability that the digits 1,2,3,4 appear as neighbours in that order, is

• A. $\frac{1}{2}$
• B. $\frac{1}{128}$
• C. $\frac{1}{256}$
• D. $\frac{1}{336}$

Answer 1

Answer: (D)

$\frac{1}{336}$

Number are $1,2,3,4,5,6,7$ and $8$

Total arrangement of those numbers are $=8!$

For $1,2,3$ and $4$ to appear as numbers

$\underset{–}{5}\underset{–}{1234}\underset{–}{6}\underset{–}{7}\underset{–}{8}$

Arrangement $=5!$

$\Rightarrow$ Probability $=\frac{Arrangementwith1,2,3,4together}{Totalarrangements}$

$=\frac{5!}{8!}$

$=\frac{5!}{8\times 7\times 6\times 5!}$

$=\frac{1}{336}$