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ashish1354
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can some one explain the solution to this problem
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Number of actual directors "lost" Sets problem
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Source: Beat The GMAT — Problem Solving |
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ashish1354
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ashish1354
- Master | Next Rank: 500 Posts
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ashish1354
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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.
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dally_gmat
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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.
