UPSC Prelims 2010 Q117: Science & Technology – Quantitative Aptitude

UPSC Prelims 2010, General Studies Paper I · Science & Technology

In how many ways can four children be made to stand in a line such that two of them, A and B, are always together?

Answer: (b) 12

Explanation

Arranging 4 children with two always together

Four children stand in a line, with A and B always adjacent to each other.

Step	Reasoning
Treat A and B as one block	This block can be arranged internally as AB or BA: 2 ways
Remaining units to arrange	The AB-block plus the other 2 children = 3 units total
Arrangements of 3 units	3! = 6 ways
Total arrangements	6 (unit arrangements) × 2 (internal AB/BA order) = 12

Why this 'block' method works:

Point	Detail
A, B fixed together	Treated as a single movable unit
Internal order	AB or BA, doubling the count

Answer: (b) 12.

(a) Incorrect. 6 would be the count WITHOUT accounting for the AB/BA internal order doubling.
(b) CORRECT. 3! (unit arrangements) × 2 (AB/BA order) = 12.
(c) Incorrect. 18 does not match the correct block-method calculation.
(d) Incorrect. 24 is the total arrangements of 4 children with NO restriction (4!), ignoring the 'always together' constraint.

Treating A and B as one block (2 internal orders: AB or BA) plus the other 2 children gives 3 units to arrange (3! = 6), so total = 6 × 2 = 12.

4 children, A and B always together: 3! × 2 = 12 arrangements.

Forgetting to multiply by 2 for the AB/BA internal ordering within the block.
Using the unrestricted 4! = 24 count, ignoring the 'always together' constraint entirely.
Miscounting the number of units to arrange after forming the block.

'Two items always together' puzzles: treat them as ONE block, arrange all units, then multiply by the block's internal orderings (usually 2).

NCERT Class XI Mathematics — Permutations and Combinations
R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)