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Home  >>  CBSE XI  >>  Math  >>  Probability
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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 ?

$\begin{array}{1 1}(A)\;\large\frac{1}{62}\\(B)\;\large\frac{1}{72}\\(C)\;\large\frac{1}{52}\\(D)\;\text{None of these}\end{array} $

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1 Answer

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  • Required probability =$\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Step 1:
Given word ALGORITHM
Number of letters in the word =9
The word is arranged randomly.
$\therefore$ Total number of outcomes n(S)=9!
Given GOR have to remain together
Thereby considering GOR as a single letter
$\therefore$ No of letter=7
Number of favorable outcomes n(E) =7!
Step 2:
$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{7!}{9!}$
$\Rightarrow \large\frac{1}{9\times 8}$
$\Rightarrow \large\frac{1}{72}$
Hence (B) is the correct answer.
answered Jul 7, 2014 by sreemathi.v
 

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