If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is ___________.

First, observe that the word ‘MOTHER’ has six distinct letters. When all six letters are rearranged, we get $$6! = 720$$ different words. These words are to be imagined as written in a dictionary (lexicographic) order, so we must count how many of those words come before the word ‘MOTHER’ itself.

Alphabetically, the six letters are ordered as $$E < H < M < O < R < T.$$ We progress through the word ‘MOTHER’ one position at a time, and at each step we count how many possible words could be formed by choosing a smaller unused letter for that position and permuting the remaining letters arbitrarily.

We denote by “count” the number of such words that appear before ‘MOTHER’.

First position: The first letter of ‘MOTHER’ is $$M.$$ Unused letters smaller than $$M$$ are $$E$$ and $$H,$$ which are two in number. Once one of these two letters is fixed, the remaining five positions can be filled in $$5!$$ ways. Using the factorial rule $$n! = n(n-1)(n-2)\dotsm 1,$$ we have $$5! = 120.$$ Hence the number of words beginning with $$E$$ or $$H$$ is $$2 \times 5! = 2 \times 120 = 240.$$ So far, $$\text{count} = 240.$$

Second position: We now fix the first letter as $$M$$ and look at the second letter, which in ‘MOTHER’ is $$O.$$ The remaining unused letters are $$E, H, O, R, T.$$ Among these, the letters alphabetically smaller than $$O$$ (while $$M$$ is already used) are $$E$$ and $$H,$$ i.e. two letters. With any one of these two as the second letter, the last four positions can be arranged in $$4! = 24$$ ways. Hence the additional words are $$2 \times 4! = 2 \times 24 = 48.$$ Adding to the previous total, $$\text{count} = 240 + 48 = 288.$$

Third position: We now have the prefix $$MO.$$ The third letter in ‘MOTHER’ is $$T.$$ The remaining unused letters are $$E, H, R, T.$$ Alphabetically smaller letters than $$T$$ within these are $$E, H, R,$$ i.e. three letters. Once any one of these three is chosen for the third position, the last three positions can be filled in $$3! = 6$$ ways. Thus the extra words contributed here are $$3 \times 3! = 3 \times 6 = 18.$$ Adding, $$\text{count} = 288 + 18 = 306.$$

Fourth position: The current prefix is $$MOT.$$ The fourth letter of ‘MOTHER’ is $$H.$$ The unused letters now are $$E, H, R.$$ The letter alphabetically smaller than $$H$$ in this set is only $$E.$$ That is just one letter. After fixing this $$E$$ as the fourth letter, the remaining two positions can be arranged in $$2! = 2$$ ways. Therefore the contribution is $$1 \times 2! = 1 \times 2 = 2.$$ Update the total: $$\text{count} = 306 + 2 = 308.$$

Fifth position: Our prefix is $$MOTH.$$ The fifth letter of ‘MOTHER’ is $$E.$$ The two unused letters are $$E$$ and $$R,$$ but $$E$$ is the smallest among them. There is no letter smaller than $$E$$ left, so no new words are added at this step.

Sixth position: The final letter $$R$$ is forced, and again no additional words arise.

Thus, exactly $$308$$ words precede ‘MOTHER’ in dictionary order. Hence the position of ‘MOTHER’ itself is $$308 + 1 = 309.$$

So, the answer is $$309$$.