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a) In how many ways can the letters in REVENUE be rearranged to form distinct "words"?

b) From the arrangements in part (a), how many have all three E’s together?

c) From the arrangements in part (a), how many have no consecutive E’s?

For (a) I got $840$ and for (b) I got $120$. Are these answers correct? I'm not sure how to do (c).

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Your answers (a) and (b) correct.

My attempt for (c): Remove all E's from REVENUE. The remaining alphabets are R,V,U,N. We can make 4!=24 words by using these 4 alphabets.

For example, RVUN is one of the possible word.

Now

"_R__V__U__N_"

If you are filling three E in any plank space yields a word which having no concecutive E's.

There are 5 ways to placing three E's. So there are 5C3=10 ways.

Hence 24×10=240 words.

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For (a) and (b), your answers, $\frac{7!}{3!}$ and $5!$, are correct.

For (c), the number of arrangements is the number of all arrangements (which you already calculated) minus the number of arrangements with $3$ consequtive E-s (which you already calculated) minus the number of arrangements with two consequtive E-s (the only one you still have to calculate).

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    $\begingroup$ Agreed. However, counting the number of arrangements with exactly two consecutive $E$s is a bit tricky. $\endgroup$ – N. F. Taussig Sep 27 '18 at 8:44

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