If the letters of the word ALGORITHM are arranged

Given word is ALGORITHM

$\Rightarrow$ Total number of letters in algorithm $=9$

$\therefore$ Total number of words $=9 !$

$\mathrm{So}, \mathrm{n}(\mathrm{S})=9 !$

If 'GOR' remain together, then we consider it as one group.

$\therefore$ Number of letters $=7$

Number of words, if ' $\mathrm{GOR}$ ' remain together in the order $=7 !$

So, $n(E)=7 !$

Required Probability $=\frac{\text { Number of favourable outcome }}{\text { Total number of outcomes }}$

$=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$

$=\frac{7 !}{9 !}[\because n !=n \times(n-1) \times(n-2) \ldots 1]$

$=\frac{7 !}{9 \times 8 \times 7 !}=\frac{1}{72}$