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Using all the letter of the word plan how numerous different words making use of all letters at a time can be made such the both A, both E, both R both N take place together .



$egingroup$ In general if you have $n$ objects through $r_1$ objects that one kind, $r_2$ objects of another,...,and $r_k$ objects the 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 an eleven-letter word.

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

See more: 8+8=0 Hence, The Sum Of Deviations About The Mean Always Equals What?

However, due to the fact that there room repeating letters, we need to divide to eliminate the duplicates accordingly.There room 2 As, 2 Rs, 2 Ns, 2 Es

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


The word setup has $11$ letters, not all of them distinct. Imagine the they are written on small Scrabble squares. And also suppose we have $11$ continually slots right into which to put these squares.

There room $dbinom112$ means to choose the slots wherein the two A"s will go. Because that each of these ways, there room $dbinom92$ means to decide where the two R"s will certainly go. For every decision about the A"s and R"s, there room $dbinom72$ ways to decide where the N"s will certainly go. Similarly, there are currently $dbinom52$ means to decide where the E"s will go. That pipeline $3$ gaps, and also $3$ singleton letters, which have the right to be i ordered it in $3!$ ways, for a total of $$inom112inom92inom72inom523!.$$


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