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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 \$\$inom112inom92inom72inom523!.\$\$

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

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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