Permutations and Combinations Doubt

Question

Question:

How many words, with or without meaning, can be made using the letters of the word DEBOTRI such that there are always two letters between D and E?

I got $4 \times 2 \times 5P_5 = 960$, but I'm not sure if it's correct.

Answer 1

Yes, you are correct!

There can be four position for fixing D or E

xooxooo oxooxoo ooxooxo oooxoox

but, D and E can also change positions so $2!$

Rest $5$ can be arranged in $^5$P$_5$ ways,

So, the total number of combinations are,

$\displaystyle{4\cdot2!\cdot^5}$P$_5$

Answer 2

There are 2! Ways to arrange D and E.

Between D and E, there are 2 positions can be filled by 5 letters (namely, B, O, T, R, I). The first can be filled in 5 ways and the second can then be filled in 4 ways.

After the above action, tie the 4 chosen objects together as one unit. Therefore, there are (3 + 1) objects to permute among themselves in 4! Ways.

“Summing” all up, we have 2!*(5*4)*4! Ways. [Same as your answer.]