CSAT Solved Papers/ 2023/Q68

2023 CSAT — Q68

Quant Counting & combinatorics 2.5 marks Medium

There are five persons $P, Q, R, S$ and $T$ each one of whom has to be assigned one task. Neither $P$ nor $Q$ can be assigned Task- $1$ . Task- $2$ must be assigned to either $R$ or $S$ . In how many ways can the assignment be done?

A 6
B 12
C 18
D 24 Answer

Worked rationale

Five persons map one-to-one onto five tasks (a bijection). Assign the most-constrained slots first.

Task- $2$ must go to $R$ or $S$ : $2$ choices.
Task- $1$ must go to someone in $\{R, S, T\}$ (not $P$ , not $Q$ ), but Task- $2$ already used one of $R, S$ . So the people still available for Task- $1$ are $\{R, S, T\}$ minus the one placed at Task- $2$ : $3 - 1 = 2$ choices.
Tasks $3, 4, 5$ : the remaining $3$ persons fill them in $3! = 6$ ways.

By the multiplication principle:

2 \times 2 \times 6 = 24.

Answer: (d) 24.

Why the other options miss

A

missed a case: assigns only the two constrained tasks ( $2\times ?$ ) and forgets the $3!$ arrangements of the remaining people.
B

an arithmetic slip: takes Task- $1$ as having only $1$ valid person (just $T$ ) instead of $2$ , halving the count ( $2\times 1\times 6$ ).
C

mis-counts the overlap: mis-handles the overlap between the Task- $1$ and Task- $2$ restrictions (e.g. $3\times 3\times 2$ ), over- or under-counting the dependency.

Specialist insight

Order the assignment by constraint tightness: fill Task- $2$ ( $2$ ways) before Task- $1$ , because Task- $2$ consumes one of $R/S$ and thereby changes Task- $1$ ’s available pool to exactly $2$ ( $\{R,S,T\}$ minus the used one). Then the unconstrained tail is $3! = 6$ . The subtle point is that the two restrictions interact through the shared people $R, S$ ; assigning the overlapping slot first keeps the count clean and avoids the $18$ trap.

The trap, in one line

Fill Task- $2$ first ( $R$ / $S$ : $2$ ), then Task- $1$ from $\{R,S,T\}$ minus that person ( $2$ ), then $3! = 6$ : $2\cdot 2\cdot 6 = 24$ $=$ (d).

← All 2023 CSAT questions