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Using all the letter of the word plan how many different words using all letters at a time deserve to be do such the both A, both E, both R both N happen together .


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$\begingroup$ In general if you have actually $n$ objects v $r_1$ objects of one kind, $r_2$ objects of another,...,and $r_k$ objects that the $k$th kind, they deserve to be arranged in $$\fracn!(r_1!)(r_2!)\dots(r_k!)$$ ways. $\endgroup$
"ARRANGEMENT" is an eleven-letter word.

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

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However, since there room repeating letters, we have to divide to eliminate the duplicates accordingly.There room 2 As, 2 Rs, 2 Ns, 2 Es

Therefore, there space $\frac11!2!\cdot2!\cdot2!\cdot2!=2494800$ ways of arranging it.


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The word arrangement has $11$ letters, not all of them distinct. Imagine the they space written on tiny Scrabble squares. And suppose we have actually $11$ continuous slots into which to put these squares.

There are $\dbinom112$ means to choose the slots whereby the two A"s will go. Because that each of this ways, there room $\dbinom92$ methods to decide wherein the two R"s will go. Because that every decision about the A"s and also R"s, there are $\dbinom72$ means to decide whereby the N"s will go. Similarly, over there are currently $\dbinom52$ methods to decide wherein the E"s will go. That pipeline $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a full of $$\binom112\binom92\binom72\binom523!.$$


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In how countless ways have the right to the letter of words 'arrange' be i ordered it if the 2 r's and also the two a's execute not happen together?
In how many ways can the letters of indigenous $PERMUTATIONS$ be i ordered it if there are constantly 4 letters in between P and also S?
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