Question:
Let \(S\) be the set of all words formed by arranging all the letters of the word HOMOGENEOUS. If a word is randomly chosen from the set \(S\), then the probability that the word selected has all the consonants together is
Let \(S\) be the set of all words formed by arranging all the letters of the word HOMOGENEOUS. If a word is randomly chosen from the set \(S\), then the probability that the word selected has all the consonants together is
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When several letters must stay together, treat them as a single block first. Arrange the block with the remaining letters and then multiply by the internal arrangements of the letters within the block.
Updated On: Jul 29, 2026
- \(\dfrac{6!\,6!}{11!}\)
- \(\dfrac{7!\,5!}{11!}\)
- \(\dfrac{6!\,5!}{11!}\)
- \(\dfrac{8!\,5!}{11!}\)
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The Correct Option is B
Solution and Explanation
Concept:
Probability
\[
=
\frac{\text{Number of favourable arrangements}}
{\text{Total number of arrangements}}.
\]
When all consonants are together, treat them as a single block and then arrange the resulting objects.
Step 1: Find the total number of arrangements of HOMOGENEOUS. The word HOMOGENEOUS contains \(11\) letters. The repeated letters are \[ O,O,O \] and \[ E,E. \] Hence, \[ n(S) = \frac{11!}{3!\,2!}. \]
Step 2: Identify vowels and consonants. Vowels: \[ O,O,O,E,E,U \] which are \(6\) letters. Consonants: \[ H,M,G,N,S \] which are \(5\) distinct letters.
Step 3: Count arrangements with all consonants together. Treat the \(5\) consonants as one block. Then we have \[ 1+6=7 \] objects: \[ [\text{Consonant Block}],\;O,\;O,\;O,\;E,\;E,\;U. \] These can be arranged in \[ \frac{7!}{3!\,2!} \] ways. The consonants inside the block can be arranged in \[ 5! \] ways. Therefore, \[ \text{Favourable arrangements} = \frac{7!}{3!\,2!}\times 5!. \]
Step 4: Calculate the probability. \[ P = \frac{\frac{7!}{3!\,2!}\times 5!} {\frac{11!}{3!\,2!}}. \] \[ = \frac{7!\,5!}{11!}. \] Therefore, \[ \boxed{\frac{7!\,5!}{11!}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Find the total number of arrangements of HOMOGENEOUS. The word HOMOGENEOUS contains \(11\) letters. The repeated letters are \[ O,O,O \] and \[ E,E. \] Hence, \[ n(S) = \frac{11!}{3!\,2!}. \]
Step 2: Identify vowels and consonants. Vowels: \[ O,O,O,E,E,U \] which are \(6\) letters. Consonants: \[ H,M,G,N,S \] which are \(5\) distinct letters.
Step 3: Count arrangements with all consonants together. Treat the \(5\) consonants as one block. Then we have \[ 1+6=7 \] objects: \[ [\text{Consonant Block}],\;O,\;O,\;O,\;E,\;E,\;U. \] These can be arranged in \[ \frac{7!}{3!\,2!} \] ways. The consonants inside the block can be arranged in \[ 5! \] ways. Therefore, \[ \text{Favourable arrangements} = \frac{7!}{3!\,2!}\times 5!. \]
Step 4: Calculate the probability. \[ P = \frac{\frac{7!}{3!\,2!}\times 5!} {\frac{11!}{3!\,2!}}. \] \[ = \frac{7!\,5!}{11!}. \] Therefore, \[ \boxed{\frac{7!\,5!}{11!}} \] \[ \boxed{\text{Answer = (B)}} \]
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