5. Venn Diagrams with 4 Sets
The following is a Venn Diagram with 4 sets - A, B, C and D. In small letters from a to p are the numbers of elements in each of the 16 different regions formed. For instance, k is the number of elements that are in sets A, B and D but not in set C.
Let P, Q, R and S be the number of elements in exactly 1 set, exactly 2 sets, exactly 3 sets and all 4 sets respectively.
n(A ∪ B ∪ C ∪ D)
= P + Q + R + S (Property 1)
n(A) + n(B) + n(C) + n(D)
=
P + 2Q + 3R + 4S (Property 2)
n(U) = P + Q + R + S + p (Property 3)
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Questions involving
4 sets, albeit rare, have been asked in the past. We use 4 rectangles - 2 horizontal and 2 vertical - to denote the 4 sets. Please note the way a 4-set Venn diagram is drawn.
Example 10
A class of 150 students were asked if they liked or disliked each of the 4 colours - red, blue, green and black. It was noted that 80,70,90 and 60 students liked the colours red, blue, green and black respectively. 30 students did not like any of the 4 colours, 40 students liked exactly 2 of the colours and 10 students liked exactly 3 of the colours. How many students liked all 4 colours?
Solution
Let Red, Blue, Green and Black be the sets comprising of students who liked these respective colours. Let P, Q, R and S be the number of students who liked exactly 1,2,3 and 4 colours respectively. Let X be the number of students who did not like any of the colours.
Total students = P + Q + R + S + X =150
⇒ P +40+10+S+30=150 ⇒ P + S =70 -----(1)
n(Red) + n(Blue) +n(Green) +n(Black) = P + 2Q + 3R + 4S
⇒ 80+70+90+60= P +80+30+4S ⇒ P +4S =190 -----(2)
Subtracting (1) from (2) ⇒ 3S = 120 ⇒ S = 40
Answer: 40
Example 11
In a club of 42 members, each of them played at least one of chess, carrom, rummy and poker. 18,22,24 and 28 members played chess, carrom, rummy and poker respectively. A total of 20 members played exactly one game and number of people who played only chess, only carrom, only rummy and only poker are in increasing order of an arithmetic progression where all the terms are even. No member plays exactly 2 games. 12 members play chess & carrom, 12 play chess & rummy, and 14 play chess & poker.
(I) How many members play all 4 games?
(II) How many members do not play chess but play carrom or rummy?
(III) How many members play chess, carrom and rummy?
Solution
Case I: 20 members play exactly 1 game and 0 members play exactly 2 games. Let P, Q, R and S be the number of people who play exactly 1, exactly 2, exactly 3 and exactly 4 games respectively.
P + Q + R + S =42 ⇒ 20+0+ R + S =42
⇒ R + S =22 -----(1)
P +2Q +3R +4S =18+22+24+28
⇒ 3R +4S =72 -----(2)
Solving (1) and (2), we get R =16 and S =6
Case II: 6 players play all 4 games. So, the centre square that is common to all 4 sets is filled with 6.
There are 6 regions that are common to exactly 2 of the 4 sets. These are to be filled with 0, as there is no member who plays
exactly 2 games.
Number of players who play exactly 1 game is 20. As the 4 regions are in AP with common difference =2, the 1st and 4th terms are a and a+6.
Average of 4 terms =2a+a+6=420 ⇒ a=2
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∴ Members playing only chess, only carrom, only rummy and only poker are
2,4,6 and 8 respectively.
Let p,q,r & s be the number of players who play exactly 3 games pertaining to the regions shown above.
From the question,
we can form the following equations.
P + Q + R + S = 16
[16 members play exactly 3 games]
P + R + 6 =12 ⇒ P + R =6
[12 play chess & carrom]
Q + R + 6 =12 ⇒ Q + R =6
[12 play chess & rummy]
P + Q + 6 =14 ⇒ P + Q =8
[14 play chess & poker]
Solving the 4 equations above,
we get P = 4, Q = 4, R = 2, S = 6
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∴ Members who do not play chess but play carrom or rummy
=6+4+6= 16
Case III: Number of members playing chess, carrom and rummy =6+2= 8
Answer: (I) 6; (II) 16; (III) 8
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