Question

In a conference there are $11$ mechanical engineers and $7$ metallurgical engineers. In how many ways can they be seated in a row such that all the metallurgical engineers do not sit together?

• A. $18!-(12!\times 7!)$
• B. ${}^{18}{P}_4-2!$
• C. ${}^{18}{P}_4\times 11$
• D. $18!-11!$

Answer 1

Answer: (A)

$18!-(12!\times 7!)$

$11$ mechanical engineers

$7$ metallurgical

concider all of them to be one unit

integral arrangements $=7!$

total arrangements such that all metallurgical engineers are together

$\Rightarrow 12!\times 17!$

Arrangements of eighteen engineers $=18!$

Arrangements such that nometallurgical engineers are together $=18!-(12!\times 7!)$