Three girls and five boys are to be seated in a row so that no two girls sit together. Total no. of ways of this arrangement are:
Options
A14,400
B120
C
D
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Permutations and CombinationsMCQPYQ Nov. 19Question 1621 of 251
All Questions Correct Answer
✅ Option a — 14,400
All Options:
- A14,400
- B120
- C
- D
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Detailed Solution & Explanation
Let us solve this problem step-by-step using the gap method:
We have 5 boys and 3 girls to be arranged in a row such that no two girls sit together.
1. **Arrange the boys:** First, arrange the 5 boys in a row. The number of ways to do this is:
2. **Identify gaps for the girls:** The 5 arranged boys create 6 gaps around them (one before the first boy, four between the boys, and one after the last boy):
\\text{\\_ } B_1 \\text{ \\_ } B_2 \\text{ \\_ } B_3 \\text{ \\_ } B_4 \\text{ \\_ } B_5 \\text{ \\_ }
3. **Arrange the girls:** The 3 girls must be placed in these 6 gaps such that there is at most one girl in each gap. The number of ways to choose and arrange the 3 girls in these 6 gaps is:
4. **Total arrangements:** By the multiplication principle, the total number of ways is:
This mathematically correct value (14,400) corresponds to Option A. However, the textbook key incorrectly lists Option B (120) as correct, which is a typographical error (likely caused by neglecting the multiplication of the two steps). We proceed with the correct mathematical derivation.
Hence, **Option A** is the correct answer.
We have 5 boys and 3 girls to be arranged in a row such that no two girls sit together.
1. **Arrange the boys:** First, arrange the 5 boys in a row. The number of ways to do this is:
2. **Identify gaps for the girls:** The 5 arranged boys create 6 gaps around them (one before the first boy, four between the boys, and one after the last boy):
\\text{\\_ } B_1 \\text{ \\_ } B_2 \\text{ \\_ } B_3 \\text{ \\_ } B_4 \\text{ \\_ } B_5 \\text{ \\_ }
3. **Arrange the girls:** The 3 girls must be placed in these 6 gaps such that there is at most one girl in each gap. The number of ways to choose and arrange the 3 girls in these 6 gaps is:
4. **Total arrangements:** By the multiplication principle, the total number of ways is:
This mathematically correct value (14,400) corresponds to Option A. However, the textbook key incorrectly lists Option B (120) as correct, which is a typographical error (likely caused by neglecting the multiplication of the two steps). We proceed with the correct mathematical derivation.
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 SyllabusExam 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.
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