Question 1149724
<pre>
ENTERTAINMENT

In alphabetical order, the letters are A,E,E,E,I,M,N,N,N,R,T,T,T.

First we find the number of distinguishable arrangements, regardless of
whether the three E's are apart or not.

That's 13 things with 3 indistinguishable E's, 3 indistinguishable N's, and 3
indistinguishable T's,

That's {{{13!/(3!3!3!)=28828800}}}

From that we will subtract the ones with some E's together. They consist of
two cases:

Case 1. the number of arrangements that have all three E's together, like EEE.
That's the number of arrangements of these 11 "things" with 3
indistinguishable N's and 3 indistinguishable T's.

A,EEE,I,M,N,N,N,R,T,T,T.

That's {{{11!/(3!3!)=1108800}}} ways.

and

Case 2. the number of arrangements that have EE together and the E apart from it.

First we find the number of arrangements without any E's.  That's the number

of arrangements of these 10 things, which have 3 indistinguishable N's and 3
indistinguishable T's:

A,I,M,N,N,N,R,T,T,T.

That's {{{10!/(3!3!)=100800}}}

Then we'll insert an EE and an E among them so that we don't put them
together, avoiding counting again the ones with EEE from case 1.

To do that, we now look at a random arrangement from the 100800 with no E's at
all, say this one:

T,R,N,A,I,T,N,N,M,T

We put 8 spaces between the letters, 1 space in the beginning, and 1 space at
the end. That's 10 spaces, and we'll put an E in one of them and EE in another
one.  That way the E and the EE won't be together.

_T_R_N_A_I_N_N_M_T_

We can choose the space to put the single E in in 10 ways.
That leaves 9 spaces to put the double EE in.

So for each of the 100800 ways we can insert the E and the EE in 10∙9 or 90 
ways.  That's 100800∙90 = 9072000 ways to have EE separate from E.

So the total number that we must subtract from 28828800 is 1108800 from case 1
and 9072000 from case 2.

28828800-1108800-9072000 = 15348000 ways.

Edwin</pre>