Question 1022046
<pre>
First we find the number of distinguishable permutations of
ARRANGEMENT.  There are 2 A's, 2 R's, 2 E's, and 2 N's. So
that's {{{11!/(2!2!2!2!)}}}{{{""=""}}}{{{2494800}}}

From that we must subtract the number of distinguishable 
permutations in which the A's, R's or both come together.

We use the "sieve" formula: 

N(X or Y) = N(X) + N(Y) - N(X and Y)
where N() means "the number of elements of". 

N(permutations with A's together OR R's together) =

N(permutations with A's together) + 

N(permutations with R's together) -

N(permutations with A's together AND R's together)

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Let's get this first: N(permutations with A's together)

They are the distinguishable arrangements of these
10 things:

(AA),R,R,N,G,E,M,E,N,T

There are 2 R's, 2 N's, 2 E's, but only one (AA)

That's {{{10!/(2!2!2!)}}}{{{""=""}}}{{{453600}}}
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Let's get this next: N(permutations with R's together)

They are the distinguishable arrangements of these
10 things:

{A,A,(RR),N,G,E,M,E,N,T}

That's also {{{10!/(2!2!2!)}}}{{{""=""}}}{{{453600}}}
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Now we get: N(permutations with A's together AND R's together)

They are the distinguishable arrangements of these
9 things:

(AA),(RR),N,G,E,M,E,N,T

There are 2 N's, and 2 E's, but only one (AA) and one (RR)

That's {{{9!/(2!2!)}}}{{{""=""}}}{{{90720}}}

So the number we must subtract from the 2494800 is

{{{453600 + 453600 - 90720}}}{{{""=""}}}{{{362880}}}

So the final answer is

{{{2494800 - 362880}}}{{{""=""}}}{{{2131920}}}

Edwin</pre>