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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?
Answers (1)
ALGORITHM... (given word)
Total number of letters: 9
Thus, total number of words = 9!
Thus, n(s) = 9!
Considering ‘GOR’ as one group
A L GOR I T H M
↓ ↓ ↓ ↓ ↓ ↓ ↓
1 2 3 4 5 6 7
Thus, no. of letters = 7
Now, if the GOR group remains together, then the order = 7!
Thus, n (E) = 7!
Now,
required probability = number of favourable outcomes / total number of outcomes
= n (E) / n(S) …..[Since n! = n x (n – 1) x (n – 2)…1]
= 7! / 9!
= 7! / 9 x 8 x 7!
= 1 / 72
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