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Using every the letter of the word arrangement how countless different words making use of all letters at a time can be do such the both A, both E, both R both N happen together .  $\begingroup$ In basic if you have $n$ objects v $r_1$ objects of one kind, $r_2$ objects that another,...,and $r_k$ objects of the $k$th kind, they have the right to be i ordered it in $$\fracn!(r_1!)(r_2!)\dots(r_k!)$$ ways. $\endgroup$
"ARRANGEMENT" is one eleven-letter word.

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

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

Therefore, there are $\frac11!2!\cdot2!\cdot2!\cdot2!=2494800$ ways of arranging it. The word plan has $11$ letters, not every one of them distinct. Imagine the they are written on little Scrabble squares. And also suppose we have actually $11$ continuous slots into which to put these squares.

There space $\dbinom112$ ways to choose the slots wherein the 2 A"s will go. Because that each of this ways, there room $\dbinom92$ means to decide where the two R"s will go. For every decision around the A"s and also R"s, there room $\dbinom72$ means to decide whereby the N"s will go. Similarly, over there are currently $\dbinom52$ ways to decide whereby the E"s will go. That pipeline $3$ gaps, and $3$ singleton letters, which have the right to be arranged in $3!$ ways, for a complete of $$\binom112\binom92\binom72\binom523!.$$ Highly energetic question. Knife 10 reputation (not counting the association bonus) in order to answer this question. The reputation requirement helps defend this concern from spam and also non-answer activity.

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In how many ways can the letter of the word 'arrange' be i ordered it if the two r's and also the two a's perform not happen together?
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