Permutations and CombinationsMCQMTP March 21Question 1720 of 249
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Find the number of combinations of the letters of the word COLLEGE taken four together:

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option a — 18

All Options:

  • A18
  • B16
  • C20
  • D26

Detailed Solution & Explanation

The word is **COLLEGE**.
Let us first analyze the letters in the word **COLLEGE**:
- Total letters = 7\displaystyle 7
- The letters and their frequencies are:
- C×1\displaystyle C\displaystyle \text{C} ×\displaystyle \times 1
- O×1\displaystyle O\displaystyle \text{O} ×\displaystyle \times 1
- L×2\displaystyle L\displaystyle \text{L} ×\displaystyle \times 2
- E×2\displaystyle E\displaystyle \text{E} ×\displaystyle \times 2
- G×1\displaystyle G\displaystyle \text{G} ×\displaystyle \times 1

There are 5\displaystyle 5 distinct letters: {C,O,L,E,G}\displaystyle \{C\displaystyle \text{C}, O\displaystyle \text{O}, L\displaystyle \text{L}, E\displaystyle \text{E}, G\displaystyle \text{G}\}. among these, the letters L\displaystyle L\displaystyle \text{L} and E\displaystyle E\displaystyle \text{E} are repeated twice, while C\displaystyle C\displaystyle \text{C}, O\displaystyle O\displaystyle \text{O}, and G\displaystyle G\displaystyle \text{G} appear once. We want to find the number of combinations (selections) of 4\displaystyle 4 letters taken together. We can classify these combinations into three mutually exclusive cases:

**Case 1: All 4 letters are distinct**
We select 4\displaystyle 4 distinct letters out of the 5\displaystyle 5 distinct letters {C,O,L,E,G}\displaystyle \{C\displaystyle \text{C}, O\displaystyle \text{O}, L\displaystyle \text{L}, E\displaystyle \text{E}, G\displaystyle \text{G}\}.
Ways1=5C4=5Ways1=5C4=5\displaystyle \text{Ways}_1 = ^{5}C_{4} = 5

**Case 2: 2 letters are alike (a pair) and 2 letters are distinct**
- First, we must select 1\displaystyle 1 pair from the 2\displaystyle 2 available repeated letters (LL\displaystyle LL\displaystyle \text{LL} or EE\displaystyle EE\displaystyle \text{EE}). The number of ways to choose this is 2C1=2\displaystyle ^{2}C_{1} = 2.
- Once the pair is chosen, the remaining 2\displaystyle 2 letters must be distinct and chosen from the remaining 4\displaystyle 4 distinct letters. The number of ways to choose them is 4C2\displaystyle ^{4}C_{2}.
Ways2=2C1×4C2=2×4×32×1=2×6=12Ways2=2C1\displaystyle \text{Ways}_2 = ^{2}C_{1} ×\displaystyle \times ^{4}C_{2} = 2 ×\displaystyle \times 4×32×1=2\displaystyle \frac{4 \times 3}{2 \times 1} = 2 ×\displaystyle \times 6 = 12

**Case 3: 2 letters are alike of one kind and 2 are alike of another kind (2 pairs)**
We must select both pairs LL\displaystyle LL\displaystyle \text{LL} and EE\displaystyle EE\displaystyle \text{EE}.
Ways3=1(i.e., the selection is {L,L,E,E})Ways3=1\displaystyle \text{Ways}_3 = 1 \quad (i.e., the selection is \displaystyle \text{i.e., the selection is } \{L\displaystyle \text{L}, L\displaystyle \text{L}, E\displaystyle \text{E}, E\displaystyle \text{E}\})

**Total number of combinations:**
Total Combinations=Ways1+Ways2+Ways3Total Combinations=Ways1+Ways2+Ways3\displaystyle \text{Total Combinations} = \text{Ways}_1 + \text{Ways}_2 + \text{Ways}_3
Total Combinations=5+12+1=18Total Combinations=5+12+1=18\displaystyle \text{Total Combinations} = 5 + 12 + 1 = 18

**Discrepancy Note:**
The mathematical analysis proves that the number of unique combinations is exactly $18\displaystyle , which corresponds to **Option A**. The textbook answer key has a typographical error, incorrectly indicating **Option C** ($20) as the correct answer.

Hence, **Option A** is the correct answer.

About This Chapter: Permutations and Combinations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Factorials, Permutations, Combinations

This chapter deals with the fundamental principles of counting. It covers factorials, circular permutations, restricted permutations, combinations, and the differences between selecting items versus arranging them.

View Official ICAI Syllabus

Exam Strategy Tip

The most common mistake is confusing 'P' (Arrangement) with 'C' (Selection). If order matters (like opening a lock), use P. If order doesn't matter (like choosing a team), use C.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

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