UPSC Prelims 2006 Q137: Science & Technology – Quantitative Aptitude
Each of 2 women and 3 men is to occupy one chair out of 8 chairs, each of which is
numbered from 1 to 8. First, women are to occupy any two chairs from those numbered 1 to 4;
and then 3 men would occupy any three chairs out of the remaining 6 chairs. What is the
maximum number of different ways in which this can be done?
- (a) 40
- (b) 132
- (c) 1440
- (d) 3660
Answer: (c) 1440
Explanation
Seating 2 women and 3 men in numbered chairs
The two women are distinct persons and the three men are distinct persons, so order of seating matters.
| Step | Calculation |
|---|---|
| Women's chairs (any two of chairs 1 to 4) | 4 x 3 = 12 ways |
| Men's chairs (any three of the remaining 6 chairs) | 6 x 5 x 4 = 120 ways |
| Total ways | 12 x 120 = 1440 |
The maximum number of different ways is 1440.
Answer: (c) 1440.
Why each option is right or wrong
- (a) Incorrect. 40 equals C(4,2) x C(6,3)/3 and ignores the order of seating persons.
- (b) Incorrect. 132 does not follow from 12 x 120.
- (c) CORRECT. 12 ways for the women and 120 for the men give 1440.
- (d) Incorrect. 3660 overcounts the placements.
Key point
The women can be seated in 4 x 3 = 12 ways and the men in 6 x 5 x 4 = 120 ways, giving 1440.
Remember
12 ways for the women x 120 ways for the men = 1440.
Common mistakes
- Using combinations instead of ordered arrangements for distinct persons.
- Forgetting that after seating the women only 6 chairs remain.
- Adding the two counts instead of multiplying them.
Exam tip
Distinct persons in numbered chairs: use permutations P(n, r), and multiply the stages.
Sources
- NCERT Class XI Mathematics — Permutations and Combinations
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)