Análise Combinatória 4: Permutações com repetição
Autor: Osmar Lewinski | Publicado: 2011-07-23T11:29:00-07:00

PERMUTAÇÃO COM REPETIÇÃOO cálculo do número de permutações com repetições é feito calculando-se o fatorial do número de elementos existentes e dividindo-se pelo fatorial dos elementos que se repetem.Ex) Quantos anagramas podem ser formados com a palavra macaco?Primeiro: note que a palavra macaco possui 6 letras.Segundo: note que a letra a repete-se 2 vezes e a letra c também repete-se 2 vezes. Dessa forma temos uma permutação de 6 elementos (já que anagramas nada mais são do que palavras com ou sem sentido formadas a partir das letras de uma palavra qualquer) em que 2 desses elementos (letras a e c) repetem-se 2 vezes.Assim temos: ¨&nbsp;&nbsp;&nbsp;&nbsp;6! =&nbsp;6.5.4.3.2.1=720=180
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2!.2!&nbsp;&nbsp;&nbsp;2.1.2.1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 4<shape equationxml="&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot; standalone=&quot;yes&quot;?&gt;
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w:h-ansi=&quot;Cambria Math&quot;/&gt;&lt;wx:font wx:val=&quot;Cambria Math&quot;/&gt;&lt;w:i/&gt;&lt;/w:rPr&gt;&lt;m:t&gt;=180&lt;/m:t&gt;&lt;/m:r&gt;&lt;/m:oMath&gt;&lt;/m:oMathPara&gt;&lt;/w:p&gt;&lt;w:sectPr wsp:rsidR=&quot;00000000&quot;&gt;&lt;w:pgSz w:w=&quot;12240&quot; w:h=&quot;15840&quot;/&gt;&lt;w:pgMar w:top=&quot;1417&quot; w:right=&quot;1701&quot; w:bottom=&quot;1417&quot; w:left=&quot;1701&quot; w:header=&quot;720&quot; w:footer=&quot;720&quot; w:gutter=&quot;0&quot;/&gt;&lt;w:cols w:space=&quot;720&quot;/&gt;&lt;/w:sectPr&gt;&lt;/w:body&gt;&lt;/w:wordDocument&gt;" id="_x0000_i1025" style="height: 25.5pt; width: 162pt;" type="#_x0000_t75">Ou seja, com as letras da palavra macaco podemos formar 180 anagramas, sendo que:ocacam e cocama são dois exemplos desses anagramas.