Worked solution · Exam P

SOA Exam P Sample Question #185 Solution

This Exam P sample reference tests Counting Techniques. The repetition condition depends only on the four-person membership and the chair; the secretary does not distinguish a state. There are C(7,4) member sets and four possible chairs per set, for 140 distinct weeks and choice B.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Counting Techniques question

Setup

Identify the information that must remain new: the four selected members together with their chair.

\text{state}=(\text{four-person set},\text{chair})

Model

Choose the four members, then choose the chair from those four.

N=\binom74(4)

Compute

Evaluate the number of distinct nonrepeating states.

\binom74=35

N=35(4)=140

Answer

A run of 140 distinct committee-chair states is possible before repetition is forced.

\boxed{140\quad\text{(B)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThe value 70 is 35(2). It allows only two chair choices within each four-person committee without justification.
CThe value 210 is 35 times C(4,2). It distinguishes an unordered pair of leaders even though repetition is defined by the chair, not by a leader pair.
DThe value 420 is 35(4)(3), which treats the secretary assignment as part of the nonrepetition condition. The same committee and chair already count as a repeat regardless of secretary.
EThe value 840 is 35(4!), which treats all four committee positions as distinct roles. Only the chair is relevant to the stated repetition.

Original practice · fully worked

Original variant: panel assignments under a pair restriction

A three-person review panel is selected from eight analysts. Two specified analysts are not allowed to serve together. After the panel is chosen, one member is assigned as lead and a different member as recorder. Calculate the number of valid panel-and-role assignments.

A
36
B
50
C
150
D
300
E
336

Answer and calculation

There are C(8,3)-6=50 valid panels after excluding those containing both specified analysts. Each valid panel has 3 choices for lead and 2 for recorder, yielding 50(6)=300 and choice D.

Setup

Count all three-person panels and subtract panels containing both restricted analysts.

\binom83=56

\text{invalid panels}=\binom61=6

Model

For each valid panel, assign the two distinct ordered roles.

\text{valid panels}=56-6=50

\text{role assignments per panel}=3(2)=6

Compute

Multiply the member-set and role-assignment counts.

50(6)=300

Answer

There are 300 valid panel-and-role assignments.

\boxed{300\quad\text{(D)}}

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How to solve this Counting Techniques question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: panel assignments under a pair restriction

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability