In how many ways can you rearrange the letters A, B, C, D, E?

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Answer 1 · 2018-07-20T01:57:53

$5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 ways$ $color(chocolate)(5! = 1 * 2 * 3 * 4 * 5 = 120 " ways"$

![http://www.statisticslectures.com/topics/permutations/]( useruploads.socratic.org )

$Formula for permutation n P r = \frac{n!}{(n - r)!}$ $"Formula for permutation " nPr = (n! )/ ((n-r)!)$

$Given n = 5, (A, B, C, D, E) & r = 5, (A, B, C, D, E)$ $"Given " n = 5, (A, B, C, D, E) " " & " " r = 5, (A, B, C, D, E)$

$:. nPr = nPn = (n!) / ((n - n)! )= (n!) / 1 = 5!$

![https://www.thoughtco.com/why-does-zero-factorial-equal-one-3126598]( useruploads.socratic.org )

![https://in.answers.yahoo.com/question/index?qid=20130723054015AAgl5E1]( useruploads.socratic.org )

$color(chocolate)(5! = 1 * 2 * 3 * 4 * 5 = 120 " ways"$

Answer 2 · 2018-07-22T12:13:51

$120$

The total number of linear arrangements obtained from $n=5$ different letters $A, B, C, D, E$ is given as

$\ ^nP_n$

$=\ ^5P_5$

$=5!$

$=5\times 4\times 3\times 2\times 1$

$=120$