Number of permutations under restrictions

Question

I need to find how many permutations are there of the letters in:

eeeddnntsc such that beside each n there is d

well i tried with a warm up saying that total permutations for the word:

$$ \left(\frac{10!}{2!*2!*3!}\right) $$

lets assume that i stick dn together, which means we now have 7 letters.

the total permutations:

$$ \left(\frac{8!*2^2}{2!*3!}\right) $$

the $2^2$ is for ordering the two pairs of dn and the 3! and 2! are for avoiding duplications.

I still however, unhappy with the outcome.

any thoughts?

much appreciation

Answer 1

There are three $e's$, two each of $d$ and $n$, and one each of $s,c,t$.

Leave out the two $d's$ for the moment

There are a total of $\dfrac{8!}{3!2!} = 3360$ permutations with $\dfrac{7!}{3!} = 840$ having the $n's$ together

In the former case, there are $2$ places for each $d$, thus $(3360-840)\times 4$ ways

In the latter case, one $d$ must be placed in between the two $n's$,
and the other has $10$ places, thus $840\times 10$ ways

Add up.

Answer 2

@true blue anil,

Thanks a lot for the explanation and solution!

would you consider my updated solution valid as well? :

$$\left(\frac{8!2^2}{2!3!}\right) +\left(\frac{2*6*6!}{3!}\right)+ \left(\frac{6*5*6!}{3!}\right) $$

explanations:

$\left(\frac{8!2^2}{2!3!}\right)$ - permutations such that dn is considered a letter and glued together.

$\left(\frac{2*6*6!}{3!}\right)$ - permutations such that ndn is considered a letter and it opens or closes the permutation. 2 stands for where ndn is , 6 - choosing where the other d is, ( must not be beside the ndn, 6! for the rest without restriction )

$\left(\frac{6*5*6!}{3!}\right)$ - permutations such that ndn is considered a letter and it isn't opening or closing the permutation. 6 stands for where ndn is, 5 - choosing where the other d is, ( must not be beside the ndn of each side, 6! for the rest without restriction )

Tom