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just wanna know what's the answer to this one.

How many ways can the word "ARRANGED" be arranged so that N and D aren't next to each other?

Thanks. :)

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    $\begingroup$ Wouldn't you rather learn how to do problems like this, instead of just knowing the answer to this one problem? $\endgroup$ – Gerry Myerson Oct 1 '15 at 9:37
  • $\begingroup$ Find in "How many ways can the word "ARRANGED" be arranged so that N and D are next to each other?" $\endgroup$ – lab bhattacharjee Oct 1 '15 at 9:41
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The total number of words is $\binom82\cdot\binom62\cdot\binom41\cdot\binom31\cdot\binom21\cdot\binom11=10080$.

The number of words with ND is $\binom72\cdot\binom52\cdot\binom31\cdot\binom21\cdot\binom11=1260$.

The number of words with DN is $\binom72\cdot\binom52\cdot\binom31\cdot\binom21\cdot\binom11=1260$.

Hence the number of words with no ND and no DN is $10080-1260-1260=7560$.

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Notice, in the given word $ARRANGED$ there are total $8$ letters out of which $A$ & $R$ are repetitive letters hence, we have

Total number of linear arrangements of $ARRANGED$ $$=\frac{8!}{2!2!}$$ Number of linear arrangements when $N$ & $D$ are kept together $$=\frac{7!}{2!2!}\cdot 2!=\frac{7!}{2!}$$ Hence, the number of linear arrangements when $N$ & $D$ are not kept together $$=\text{(total number of linear arrangements)}-\text{(number of linear arrangements when N & D are kept together)}$$ $$=\frac{8!}{2!2!}-\frac{7!}{2!}=\frac{3\cdot 7!}{2!}=\color{red}{7560}$$

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