Questions

In how many ways the letters of the word ENGINEERING, can be arranged such that all E’s come together but no two N’s come together

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\frac{5}{4}

7!

\frac{5}{4} \times 7!

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detailed solution

Correct option is C

Given the word ENGINEERING

There are 3 E’s, 3 N’s, 2 G’s, 2 I’s, 1 R

$\times E E E \times G \times G \times I \times I \times R \times \to \frac{6!}{{(2!)}^{2}}$

In 7 gaps 3 N’s can be arranged in $\frac{^{7} P_{3}}{3!}$ ways

Hence required ways are $\frac{6!}{{(2!)}^{2}} \times \frac{^{7} P_{3}}{3!}$

$\Rightarrow \frac{5}{4} \times 7!$

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