- ashish1354
- Master | Next Rank: 500 Posts
- Posts: 100
- Joined: Wed Jul 30, 2008 9:52 am
- Thanked: 4 times
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Number of actual directors "lost" Sets problem
Source: Beat The GMAT — Problem Solving |
Okay .. did i read it wrong? .. does the question mean that how many are distinct in total? then it is definitely 13.
i could'nt understand how are 13 people distinct. Please explain if you got it.
if you add all the numbers from the latest figure, 2+2+2+1+1+1+4 =13
i.e. there are 13 distint persons satisfying all the conditions and serving on the 3 boards:
condition 1. 4 serve on 3 boards each (lets say, board a, b and c)
condition 2. each pair i.e. ab, bc, ca has 5 in common, and we have 4 from condition 1. so we need 1 person common between each pair to make it total 5.
condition 3. since each board has 8 persons in total, hence there must be 2 persons on each board who work only on that board.
hence 13 different persons in total are serving on 3 boards under above conditions.
i.e. there are 13 distint persons satisfying all the conditions and serving on the 3 boards:
condition 1. 4 serve on 3 boards each (lets say, board a, b and c)
condition 2. each pair i.e. ab, bc, ca has 5 in common, and we have 4 from condition 1. so we need 1 person common between each pair to make it total 5.
condition 3. since each board has 8 persons in total, hence there must be 2 persons on each board who work only on that board.
hence 13 different persons in total are serving on 3 boards under above conditions.
If we need to apply following equation here...how do we apply here??
For 3 sets A, B, and C: P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
I am not able to understand how we can make use of this equation?
Thanks in advance for your reply and time..
For 3 sets A, B, and C: P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
I am not able to understand how we can make use of this equation?
Thanks in advance for your reply and time..
raijonney wrote:if you add all the numbers from the latest figure, 2+2+2+1+1+1+4 =13
conditions.
In terms of :
P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
P(A), P(B),P(C) is the number of people in each board i.e. 8
P(AnB), P(AnC), P(BnC) is intersection i.e. common person in 2 boards i.e. 5
P(AnBnC) is intersection of all boards, i.e. common person in all 3 boards i.e. 4
hence the equation will be => 8 + 8 + 8 - 5 - 5 - 5 + 4 = 13, hence the answer.
P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
P(A), P(B),P(C) is the number of people in each board i.e. 8
P(AnB), P(AnC), P(BnC) is intersection i.e. common person in 2 boards i.e. 5
P(AnBnC) is intersection of all boards, i.e. common person in all 3 boards i.e. 4
hence the equation will be => 8 + 8 + 8 - 5 - 5 - 5 + 4 = 13, hence the answer.
















