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Using all the letters of the word ARRANGEMENT exactly how many various words making use of all letters at a time can be made such that both A, both E, both R both N occur together .


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$egingroup$ In general if you have actually $n$ objects through $r_1$ objects of one sort, $r_2$ objects of one more,...,and also $r_k$ objects of the $k$th sort, they can be arranged in $$fracn!(r_1!)(r_2!)dots(r_k!)$$ means. $endgroup$
"ARRANGEMENT" is an eleven-letter word.

If there were no repeating letters, the answer would simply be $11!=39916800$.

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However before, because there are repeating letters, we have to divide to remove the duplicates as necessary.Tright here are 2 As, 2 Rs, 2 Ns, 2 Es

Because of this, tright here are $frac11!2!cdot2!cdot2!cdot2!=2494800$ ways of arranging it.


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Words ARRANGEMENT has $11$ letters, not every one of them distinct. Imagine that they are written on bit Scrabble squares. And mean we have actually $11$ consecutive slots right into which to put these squares.

There are $dbinom112$ ways to choose the slots where the 2 A"s will certainly go. For each of these methods, there are $dbinom92$ means to decide wbelow the 2 R"s will certainly go. For eincredibly decision about the A"s and also R"s, tbelow are $dbinom72$ methods to decide wbelow the N"s will certainly go. Similarly, tright here are now $dbinom52$ means to decide wright here the E"s will certainly go. That leaves $3$ gaps, and $3$ singleton letters, which have the right to be arranged in $3!$ means, for a total of $$inom112inom92inom72inom523!.$$


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In just how many kind of means have the right to the letters of the word 'arrange' be arranged if the 2 r's and also the 2 a's carry out not happen together?
In exactly how many type of methods have the right to the letters of word $PERMUTATIONS$ be arranged if tbelow are always 4 letters in between P and S?
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