Q.58 Using the letters in the word TRICK a new word containing five distinct letters is formed such that T appears in the middle. The number of distinct arrangements is _______.
The word TRICK has five distinct letters: T, R, I, C, K. With T fixed in the middle position of a new five-letter word, the remaining four positions must be filled using the other four distinct letters (R, I, C, K) without repetition, yielding 24 distinct arrangements.
Step-by-Step Solution
Fix T in the third (middle) position, leaving positions 1, 2, 4, and 5 to arrange R, I, C, K.
-
Position 1: 4 choices (R, I, C, or K).
-
Position 2: 3 remaining choices.
-
Position 4: 2 remaining choices.
-
Position 5: 1 remaining choice.
Total: 4×3×2×1=24. This is 4! or P(4,4)=24.
Option Analysis
No explicit options are provided, but common multiple-choice variants from similar problems include 6, 24, 72, and 120.
| Option | Calculation Error | Why Incorrect |
|---|---|---|
| 6 | 3! | Ignores two outer positions; assumes only three spots. |
| 24 | 4! | Correct: Full permutation of four letters in four spots. |
| 72 | 4!×3 | Overcounts by multiplying extra factor; no basis for it. |
| 120 | 5! | Treats all five letters freely; ignores T fixed in middle. |
Introduction to TRICK Word Arrangements T Middle
In permutation problems like TRICK word arrangements T middle, students preparing for CSIR NET or competitive exams calculate distinct five-letter words from T, R, I, C, K with T fixed centrally. This tests fundamental permutation principles without repetition, resulting in exactly 24 valid arrangements.
Core Permutation Concept
Permutations count ordered arrangements. For TRICK word arrangements T middle:
-
Total letters: 5 distinct.
-
Middle (position 3) fixed as T: 1 way.
-
Remaining 4 letters permute in 4 positions: P(4,4)=4!=24.
Formula: n!/(n−r)! where n=4, r=4. Avoids combinations since order matters.
Common Pitfalls in Calculations
-
Mistaking for combinations (C(4,4)=1): Ignores order.
-
Forgetting no repetition: Leads to 54=625 (wrong).
-
Free arrangement without fix: 5!=120 overcounts.
Practice Examples
-
Swap condition to C middle: Still 24 (same logic).
-
Three-letter subset: P(4,3)=24.
-
With repetition: 44=256 (but problem specifies distinct).
Master TRICK word arrangements T middle for exam success—practice yields precision in permutations.
