Out of 2000 employees in a office 48 % preferred cofee, 54% liked tea , 64% used to smoke . Out of th etotal employee 28% used cofee and tea , 32% used tea and smoke and 30% prefered cofee and smoke and only 6% did none of these . The number having all th ethrree is ?

Let us define the following sets

C: People like to have coffee

T: People like to have tea

S: People like to have smoke

Total employees = 2000

So,

n(C) = 48% of 2000 = 960

n(T) = 54% of 2000 = 1080

n(S) = 64% of 2000 = 1280

n(C ∩ T) = 28% of 2000 = 560

n(T ∩ S) = 32% of 2000 = 640

n(C ∩ S) = 30% of 2000 = 600

n(C ∩ S ∪ T) = 6% of 2000 = 120

n(C ∪ S ∪ T) = 2000 – 120 = 1880

Using the formula,

n(C ∪ S ∪ T) = n(C) + n(S) + n(T) – n(C ∩ S) – n(S ∩ T) – n(C ∩ T) + n(C ∩ S ∩ T)

⇒ 1880 = 960 + 1280 + 1080 - 600 – 640 – 560 + n(C ∩ S ∩ T) 

n(C ∩ S ∩ T) = 360

Hence, number of people having all three is 360.

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