This Exam P sample reference tests Counting Methods. The favorable assignments are the derangements of four labeled objects. Inclusion-exclusion gives nine derangements among 24 permutations, so the probability is 9/24=0.375 and choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 1/24=0.0417 counts one particular no-match assignment. There are nine favorable permutations.
BThe value 6/24=0.250 counts only the six four-cycles. It omits the three derangements made of two disjoint swaps.
CThe value 8/24=0.333 omits one of the nine derangements, commonly by overlooking a valid two-swap pairing.
EThe value 0.500 assumes matches and nonmatches split the permutation space evenly. Direct counting gives fifteen assignments with at least one fixed point and nine without.
Original practice · fully worked
Original variant: avoid two forbidden adjacencies
Six distinct diagnostic steps are arranged uniformly at random in a line. Step A may not be adjacent to step B, and step C may not be adjacent to step D. Calculate the probability that neither forbidden adjacency occurs.
A 2/15
B 1/3
C 7/15
D 8/15
E 2/3
Variant answer in brief
Each specified adjacent pair forms a two-order block in 2(5!) arrangements, and both blocks occur in 4(4!) arrangements. Inclusion-exclusion leaves 336 of 720 orders, giving 7/15 and choice C.
Setup
Setup
Count all linear orders of the six labeled steps.
Nall=6!=720
Model
Model
Treat each forbidden adjacency as a block. Either single block has two internal orders, while two simultaneous blocks have four.
NAB=2(5!)=240,NCD=2(5!)=240
NAB∩CD=22(4!)=96
Compute
Compute
Use inclusion-exclusion to remove orders containing either forbidden block.
Nneither=720−240−240+96=336
Pr(neither adjacency)=720336=157
Answer
Answer
Neither forbidden pair is adjacent with probability 7/15.
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