# Number of arrangements of INDEPENDENCE where 2 letters are between the $D's$

In how many arrangements between the letters of the word $\text{INDEPENDENCE}$ are there always 2 letters between the 2 $D's$?

 Unfortunately, I have no idea as to how to solve this problem, else I would certainly have included my attempts on this. I have just started learning Permutations and Combinations and am quite confused with this problem. Any help with this question would be really appreciated. Many thanks in anticipation!

PS The given answer is $226800$.

• First arrange INEPENENCE, and then put the two D's in one of the 9 possible positions. – hmakholm left over Monica Sep 18 '15 at 15:09
• Sir, the number of permutations of INEPENENCE would be $$\dfrac{10!}{3!\times 4!}$$ How should I proceed after that, Sir? – Ishan Sep 18 '15 at 15:11
• Please could you show me how to 'put the D's'? I wasn't able to understand. Also could you explain how there are 9 positions where the D's could be kept? Many thanks. – Ishan Sep 18 '15 at 15:16
• x @BetterWorld: I don't like the tone you're taking with me. Good luck finding someone who's willing to help you with that attitude. – hmakholm left over Monica Sep 18 '15 at 15:17
• @HenningMakholm I'm sorry. I call everybody Sir out of respect. I'll stop at once. Sorry for having offended you. – Ishan Sep 18 '15 at 15:17

To get an arrangement of the desired form, you can permute INEPENENCE (25200 ways to do this, as you've computed), then insert the first (leftmost) D into the permuted INEPENENCE. There are 9 ways to insert the first D--before the first letter, or the second, or ... or the ninth; if you insert the first D anywhere else, you cannot satisfy the requirement of having two letters between the D's. Then there is no choice where the second D must go. So the total is $25200\times9=226800$.
$\dfrac{10!}{3!\times 4!}$ 10 - letter perms without the Ds
so total perms = $9 \times\dfrac{10!}{3!\times 4!}=226800$