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I have clear understanding of permutation but I still can not intuitively understand why in permutation for non-distinct elements we divide number of possible arrangements by the factor of number of non-distinct elements. $n!/(n_1! \times n_2! \times \cdots\times n_k!)$

Michael Hardy
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R.Temur
  • 415

1 Answers1

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For every permutation of $AAABC$ there are six permutations of $A_1 A_2 A_3 BC,$ because $6=3!$ is the number of permutations of $A_1 A_2 A_3:$

$$ \begin{array}{c} \left. \begin{array}{l} A_1 A_2 A_3 BC \\ A_1 A_3 A_2 BC \\ A_2 A_1 A_3 BC \\ A_2 A_3 A_1 BC \\ A_3 A_1 A_2 BC \\ A_3 A_2 A_1 BC \end{array} \right\} AAABC \\[10pt] \left. \begin{array}{l} A_1 A_2 A_3 CB \\ A_1 A_3 A_2 CB \\ A_2 A_1 A_3 CB \\ A_2 A_3 A_1 CB \\ A_3 A_1 A_2 CB \\ A_3 A_2 A_1 CB \end{array} \right\} AAACB \\[10pt] \left. \begin{array}{l} A_1 A_2 B A_3 C \\ A_1 A_3 B A_2 C \\ A_2 A_1 B A_3 C \\ A_2 A_3 B A_1 C \\ A_3 A_1 B A_2 C \\ A_3 A_2 B A_1 C \end{array} \right\} AABAC \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ \text{etc.} \end{array} $$

Michael Hardy
  • 1