If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

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CBSE | Class 11 | Maths | NCERT Exemplar | Probability

If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

Solution:
The given word is ALGORITHM
$\Rightarrow$ The total no. of letters in algorithm $=9$
$\therefore$ The total no. of words $=9 !$
Therefore, $n(S)=9 !$
If ‘GOR’ remain together, then it will be considered as one group.
$\therefore$ No. of letters $=7$
No. of words, if ‘GOR’ remains together in the order = $7 !$
Therefore, $n(E)=7 !$
$\begin{array}{l} \text { Required Probability }=\frac{\text { Number of favourable outcome }}{\text { Total number of outcomes }} \\ =\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})} \\ =\frac{7 !}{9 !}[\because \mathrm{n} !=\mathrm{n} \times(\mathrm{n}-1) \times(\mathrm{n}-2) \ldots 1] \\ =\frac{7 !}{9 \times 8 \times 7 !}=\frac{1}{72} \end{array}$