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Using every the letters of the word arrangement how countless different words using all letter at a time deserve to be make such the both A, both E, both R both N take place together .
$egingroup$ In general if you have actually $n$ objects v $r_1$ objects that one kind, $r_2$ objects the another,...,and $r_k$ objects the 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 to be no repeating letters, the answer would just be $11!=39916800$.
See more: What Does The Suffix Ment Mean, Ment Meaning
However, because there are repeating letters, we have to divide to remove the duplicates accordingly.There room 2 As, 2 Rs, 2 Ns, 2 Es
Therefore, there are $frac11!2!cdot2!cdot2!cdot2!=2494800$ ways of arranging it.
The word arrangement has $11$ letters, not all of them distinct. Imagine that they room written on little Scrabble squares. And also suppose we have $11$ consecutive slots into which to placed these squares.
There room $dbinom112$ methods to choose the slots where the two A"s will go. Because that each of these ways, there room $dbinom92$ methods to decide wherein the 2 R"s will go. For every decision about the A"s and also R"s, there room $dbinom72$ methods to decide wherein the N"s will go. Similarly, there are now $dbinom52$ means to decide where the E"s will go. That pipeline $3$ gaps, and $3$ singleton letters, which have the right to be i ordered it in $3!$ ways, because that a total of $$inom112inom92inom72inom523!.$$
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In how many ways have the right to the letters of words 'arrange' be i ordered it if the two r's and the two a's carry out not take place together?
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