Arranging letters with two letters not next to each other

Question

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. :)

Answer 1

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 $$=\text{(total number of linear arrangements)}-\text{(number of linear arrangements when N & D are kept together)}$$ $$=\frac{8!}{2!2!}-\frac{7!}{2!}=\frac{3\cdot 7!}{2!}=\color{red}{7560}$$

Answer 2

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 $10080-1260-1260=7560$.