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

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