Question

Number of all possible words (with or without meaning) that can be formed using all the letters of the word CABINET in which neither the word CAB nor the word NET appear is

• A. 5040
• B. 4806
• C. 4800
• D. 5034

Hint:

For problems involving restrictions like "these letters must be together" or "this pattern must not appear", the inclusion-exclusion principle is a powerful tool. Remember the formula for two sets: Total - (A or B) = Total - (A + B - A and B). When letters must be together, treat them as a single block for permutation purposes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem requires the use of the Principle of Inclusion-Exclusion. We first find the total number of permutations of the letters. Then we find the number of permutations where "CAB" appears and where "NET" appears. We subtract these from the total, and add back the cases where both appear, as they have been subtracted twice.
Step 2: Key Formula or Approach
Let $U$ be the set of all possible permutations.
Let $A$ be the set of permutations where "CAB" appears.
Let $B$ be the set of permutations where "NET" appears.
We want to find the number of permutations where neither "CAB" nor "NET" appears, which is given by $|U| - |A \cup B|$.
Using the Principle of Inclusion-Exclusion:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
So, the required number is $|U| - (|A| + |B| - |A \cap B|)$.
Step 3: Detailed Explanation
The word is CABINET. It has 7 distinct letters: C, A, B, I, N, E, T.
Total permutations ($|U|$):
The total number of words that can be formed using these 7 distinct letters is: \[ |U| = 7! = 5040 \]
Permutations with "CAB" ($|A|$):
To count the words containing "CAB", we treat "CAB" as a single block. Now we are arranging the items: (CAB), I, N, E, T. These are 5 distinct items, so the number of arrangements is: \[ |A| = 5! = 120 \]
Permutations with "NET" ($|B|$):
Similarly, to count the words containing "NET", we treat "NET" as a single block. We are arranging the items: (NET), C, A, B, I. These are 5 distinct items, so the number of arrangements is: \[ |B| = 5! = 120 \]
Permutations with both "CAB" and "NET" ($|A \cap B|$):
To count words containing both "CAB" and "NET", we treat "CAB" as one block and "NET" as another block. We are arranging the items: (CAB), (NET), I.
These are 3 distinct items, so the number of arrangements is: \[ |A \cap B| = 3! = 6 \]
Calculate the required number:
First, find the number of words where at least one of "CAB" or "NET" appears:
\[ |A \cup B| = |A| + |B| - |A \cap B| = 120 + 120 - 6 = 234 \] The number of words where neither "CAB" nor "NET" appears is:
\[ |U| - |A \cup B| = 5040 - 234 = 4806 \]
Step 4: Final Answer
The number of possible words in which neither "CAB" nor "NET" appears is 4806.