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

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

A
$\frac{5}{4}$
B
7!
C
$\frac{5}{4} \times 7!$
D
None

Solution:

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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