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Using all the letters of the word setup how numerous different words utilizing all letters at a time deserve to be made such the both A, both E, both R both N happen together .

$egingroup$ In general if you have $n$ objects with $r_1$ objects that one kind, $r_2$ objects that 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 one eleven-letter word.

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

See more: Opposition To Current Flow Is Called, Combinations

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

Therefore, there room $frac11!2!cdot2!cdot2!cdot2!=2494800$ methods of arranging it.

The word plan has $11$ letters, not all of them distinct. Imagine that they space written on tiny Scrabble squares. And also suppose we have $11$ consecutive slots right into which to placed these squares.

There room $dbinom112$ means to **choose** the slots whereby the 2 A"s will go. Because that **each** of this ways, there are $dbinom92$ ways to decide wherein the 2 R"s will certainly go. For every decision around the A"s and R"s, there room $dbinom72$ ways to decide where the N"s will go. Similarly, there are now $dbinom52$ ways to decide where the E"s will go. That pipeline $3$ gaps, and $3$ singleton letters, which can be i ordered it in $3!$ ways, because that a total of $$inom112inom92inom72inom523!.$$

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## Not the prize you're feather for? Browse various other questions tagged permutations or asking your very own question.

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