How many different ways are there of arranging the letters in the word ACCOMMODATION if no two Cs may be together?

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Answer 1 · 2018-02-09T15:56:04

see a step process below;

Firstly you have to find the total arrangement which is;

$ACCOMMODATION = (13!)$ $"ACCOMMODATION" = (13!)$

$2C's$ $"2C's"$

$2M's$ $"2M's"$

$2A's$ $"2A's"$

$3O's$ $"3O's"$

$\frac{13!}{2! \times 2! \times 2! \times 3!} = 129729600 w a y s$ $(13!)/(2!×2!×2!×3!) =129729600 ways$

Then you have to find the arrangement if $2Cs$ $"2Cs"$ are adjacent;

$ACOMMODATION = ((13 - 1)! = 12!)$ $"ACOMMODATION" = ((13 - 1)! = 12!)$

$1C's$ $"1C's"$

$2M's$ $"2M's"$

$2A's$ $"2A's"$

$3O's$ $"3O's"$

$\frac{(13 - 1)!}{3! \times 2! \times 2!}$ $((13 - 1)!)/(3!×2!×2!)$

$\frac{12!}{3! \times 2! \times 2!} = 19958400 w a y s$ $(12!)/(3!×2!×2!) = 19958400 ways$

Then subtract both answers;

$129729600 - 19958400 = 109771200 w a y s$ $129729600 - 19958400 = 109771200ways$

Hope this helps!