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How can I find the number of ways in which the letters $$z,z,y,y,x,x,w,w,v,v$$ can be arranged so that two letters of the same kind never appear consecutively.
I am not confident that my approach is correct in regards to the initial logic to solve the problem as well as possibly over/undercounting throughout.

My answer: Logic:
Answer =
Total possible choices -
(total choices where 5 doubles occur) -
(total choices where 4 doubles occur) -
(total choices where 3 doubles occur) -
(total choices where 2 doubles occur) -
(total choices where 1 double occurs)

Total Possible Choices $$=10!$$ $$=3628800$$

Total Possible Choices Where 5 "Doubles" Occur
Let each double be equal to a singluar value - eg: z = z,z; v = v,v
This simplifys the equation to find the number of possible choices to: $$\binom{5}{1}\times\binom{4}{1}\times\binom{3}{1}\times\binom{2}{1}\times\binom{1}{1}$$ $$ =5!$$ $$=120$$ as all 10 positions will be taken up if 5 doubles are present.

Total Possible Choices Where 4 "Doubles" Occur
Not sure if this is correct $$=P(n,r)$$ $$=P(10,8)$$ $$=1814400$$

This process is followed for the remaining occurances.
3 Doubles $$=P(10,6) = 151200$$ 2 Doubles $$=P(10,4) = 5040$$ 1 Double $$=P(10,2) = 90$$

Finding the Answer $$ = 3628800 - (120) - (1814400)-(151200)-(5040)-(90)$$ $$ = 1657950$$

Is this approach remotely correct?, if not what should I be doing instead?

Dan
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  • See here: http://math.stackexchange.com/questions/580435/number-of-2n-letter-words-using-double-n-letter-alphabet-without-consecutiv –  Aug 30 '15 at 01:21

2 Answers2

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You can use Inclusion-Exclusion to find this:

Let S be the set of all arrangements of the letters, and let $A_i$ be the set of arrangements with the ith letters together, where $1\le i\le5$.

Then $\displaystyle\big|\overline{A_1}\cap\cdots\cap\overline{A_5}\big|=\big|S\big|-\sum_{i}\big|A_i\big|+\sum_{i<j}\big|A_i\cap A_j\big|-\sum_{i<j<k}\big|A_i\cap A_j\cap A_k\big|+\cdots$

$\displaystyle\hspace{1.3 in}=\frac{10!}{(2!)^5}-\binom{5}{1}\frac{9!}{(2!)^4}+\binom{5}{2}\frac{8!}{(2!)^3}-\binom{5}{3}\frac{7!}{(2!)^2}+\binom{5}{4}\frac{6!}{(2!)^1}-5!$

$\displaystyle\hspace{1.3 in}=113,400-113,400+50,400-12,600+1800-120=39, 480$

user84413
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Using the formula in my answer here, I calculate that $39480$ of the ${10\choose 2,2,2,2,2}=113400$ possible permutations have no consecutive letters the same.

Community
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