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cejismundo
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of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 21 Jul 2006, 16:00
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of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 belong to club Z. If 2 belong to all three organizations and 3 belong to exactly two organizations, how many belong to none of these organizations?

a. 1
b. 2
c. 4
d. 6
e. 11



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D

A+B+C-AB-BC-CA+ABC + NONE = TOTAL
Given
A = 5, B = 7 , C = 9
ABC = 2
AB+BC+CA-3ABC = 3 i.e AB+BC+CA - 3 * 2 = 3 i.e AB+BC+CA = 9

Put given values in above equation
5+7+9-9+2+NONE = 20
NONE = 6
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cejismundo
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I dont understand how you get
AB+BC+CA-3ABC = 3 i.e AB+BC+CA - 3 * 2 = 3 i.e AB+BC+CA = 9

can you please explain?
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ps_dahiya
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of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 11:30
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cejismundo
I dont understand how you get
AB+BC+CA-3ABC = 3 i.e AB+BC+CA - 3 * 2 = 3 i.e AB+BC+CA = 9

can you please explain?
Show more

AB - Includes that are in A as well as B and also include ABC
BC - Includes that are in B as well as C and also include ABC
CA - Includes that are in C as well as A and also include ABC

AB + BC + CA includes ABC three times. So get that are part of exactly two we need to subtract ABC three times.
Make venn diagram and it should be clear.

Hope this helps.
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Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6
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ps_dahiya
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freetheking
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6
Show more


Where it is written that n(X&Y) = 3 and n(Y&Z) = 3 and n(Z&X) =3????
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ps_dahiya
freetheking
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6

Where it is written that n(X&Y) = 3 and n(Y&Z) = 3 and n(Z&X) =3????
Show more


I was just assuming one possible way...
as long as (n(X&Y)+n(Y&Z)+n(Z&X)) = 9, it won't have any effect.
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ps_dahiya
freetheking
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6

Where it is written that n(X&Y) = 3 and n(Y&Z) = 3 and n(Z&X) =3????

I was just assuming one possible way...
as long as (n(X&Y)+n(Y&Z)+n(Z&X)) = 9, it won't have any effect.
Show more


But here we have number like this. If we modify the numbers then certainly it will be wrong.
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Some basics first.....

Let U be the universal set. Let A, B, C be member sets within U.

Place a venn diagram in front of yourself with intersecting sets A, B,C within the universal set U. Lets find an equation that must hold true.

U = a - ab - ca + abc (part of set A not intersecting with anything else)
+b - bc - ca + abc (part of set B not intersecting with anything else)
+c - ca - bc + abc ((part of set C not intersecting with anything else)
+ ab - abc (part common to A and B but not C)
+ bc - abc (part common to B and C but not A)
+ ca - abc (part common to C and A but not B)
+ abc (part common to all 3 sets)
+ C(a + b + c) (complement of union of A, B, C)

Which simplifies to

U = a+b+c-(ab+bc+ca)+abc-C(a+b+c)

This formula above holds true unconditionally. In this problem, the trick lies in plugging in the values for (ab+bc+ca). Elements that are members of 2 sets would fall into (AB+BC+CA) but there is a catch. (AB+BC+CA) includes ABC 3 times. So, number of members of exactly 2 groups
= ab - abc + bc -abc + ca - abc.

The other elements can be readily plugged in as explained above, so I won't go into that.

Note that similarly, for 2 sets, the rule is:

U = a + b + C(a+b) -ab

Hope this helps



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