In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows:

French = 17, English = 13, Sanskrit = 15

French and English = 09, English and Sanskrit = 4

French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study

(i) French only

(ii) English only

(iii) Sanskrit only

(iv) English and Sanskrit but not French

(v) French and Sanskrit but not English

(vi) French and English but not Sanskrit

(vii) at least one of the three languages

(viii) none of the three languages

Given:

Total number of students = 50

Number of students studying English = 13

Number of students studying French = 17

Number of students studying Sanskrit = 15

Number of students studying English and French = 9

Number of students studying French and Sanskrit = 5

Number of students studying English and Sanskrit = 4

Number of students studying all three subjects = 3

Let U be the total number of students, E, F and S be the number of students passed in English, French and Sanskrit respectively

n(F ∩ S ∩ E) = a = 3

n(F ∩ S) = a + d = 5

⇒ 3 + d = 5

⇒ d = 2

n(F ∩ E) = a + b = 9

⇒ 3 + b = 9

⇒ b = 6

n(S ∩ E) = a + c = 4

⇒ 3 + c = 4

⇒ c = 1

n(F) = e + d + a + b = 17

⇒ e + 2 + 3 + 6 = 17

⇒ e + 11 = 17

⇒ e = 6

n(E) = g + c + a + b = 13

⇒ g + 1 + 3 + 6 = 13

⇒ g + 10 = 13

⇒ g = 3

n(S) = f + c + a + d = 15

⇒ f + 1 + 3 + 2 = 15

⇒ f + 6 = 15

⇒ f = 9

(i) Number of students study French only = e = 6

(ii) Number of students study English only = g = 3

(iii) Number of students study Sanskrit only = f = 9

(iv) Number of students studying English and Sanskrit but not French = c = 1

(v) Number of students studying French and Sanskrit but not English = d = 2

(vi) Number of students studying French and English but not Sanskrit = b = 6

(vii) Number of students studying at least one of the three languages

= a + b + c + d + e + f + g

= 3 + 6 + 1 + 2 + 6 + 9 + 3

= 30

(viii) Number of students studying none of the three languages = Total – (a + b + c + d + e + f + g)

= 50 – (3 + 6 + 1 + 2 + 6 + 9 + 3)

= 50 – 30

= 20