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

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In how many ways can the letters of the word 'arrange' be i ordered it if the two r's and the 2 a's do not occur together?
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