Permutations and CombinationsMCQPYQ June 22Question 1618 of 251
All Questions

If four words are taken with or without meaning from the word 'LOGARITHAM' without repetition. How many words will be formed?

Options

A5040
B2520
C120
D40320
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 2520

All Options:

  • A5040
  • B2520
  • C120
  • D40320

Ad

Detailed Solution & Explanation

The word **LOGARITHAM** has 10 letters: L, O, G, A, R, I, T, H, A, M (with 'A' appearing 2 times).

If all 10 letters were distinct (like in the standard word "LOGARITHMS"), the number of 4-letter words would be 10P4=5040\displaystyle ^{10}P_4 = 5040.
Since 'A' is repeated twice, the textbook answer key attempts to adjust for this repetition by dividing the total permutations by 2!\displaystyle 2!:
textWays(asperanswerkey)=frac10P42!=frac50402=2520\\text{Ways (as per answer key)} = \\frac{^{10}P_4}{2!} = \\frac{5040}{2} = 2520
This matches Option B.

Let us also show the mathematically rigorous derivation of this problem:
The distinct letters are {L,O,G,A,R,I,T,H,M}\displaystyle \{L, O, G, A, R, I, T, H, M\} (9 distinct, with 'A' having frequency 2, and others frequency 1).
- **Case 1: All 4 letters are distinct.** We select 4 distinct letters from 9: 9P4=3024\displaystyle ^9P_4 = 3024.
- **Case 2: 2 letters are 'A' and the other 2 are distinct.** We select 2 other distinct letters from 8: 8C2timesfrac4!2!=28times12=336\displaystyle ^8C_2 \\times \\frac{4!}{2!} = 28 \\times 12 = 336.
- **Total rigorous ways:** 3024+336=3360\displaystyle 3024 + 336 = 3360.
Since 3360 is not in the options, Option B (2520) is the intended textbook answer based on the simplified division logic.
Hence, **Option B** 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

More Questions from Permutations and Combinations

The value of N\displaystyle N in N!+17!=18!\displaystyle N! + \frac{1}{7!} = \frac{1}{8!} is

A person can go from place 'A' to 'B' by 11 different modes of transport but is allowed to return to 'A' by any mode other than the one earlier. The number of different ways in which the entire journey can be completed is:

If a man travels from place A to B in 10 ways then by how many ways can he come back by another train?

If n!10(n−1)!=(n−1)!(n−3)!\displaystyle \frac{n!}{10(n-1)!} = \frac{(n-1)!}{(n-3)!} find 'n'.

Which of the following is a correct statement.

nP3=2⋅nP2\displaystyle ^nP_3 = 2 \cdot ^nP_2. Find n\displaystyle n.

Ready to Master Permutations and Combinations?

Practice all 251 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free