Question

A survey of 100 students was taken concerning their knowledge of foreign languages. The results were as follows:

30 know French
19 know German
16 know Russian
11 know both French and Russian
12 know both French and German
6 know both German and Russian
4 know all three languages

How many students do NOT know any of the three languages?

Answer 1

get the total

30 + 19 + 16 + 11 + 12 + 6 + 4

then subtract the sum from 100

got it?

Answer 2

let F be the set of students who know french
let G be the set of students who know german
let R be the set of students who know Russian
let A be the set of students who know any of the three languages
C the classroom ! All students !
It's obvious that C=AUFUGUR
and a\[\left| F \cap G \cap R\right|= 4 \]lso \[A \cap (F \cup G \cup R)= \emptyset \]
so \[\left| C \right|=\left| A \right|+\left| F \cup G \cup R\right|\] (**)(loking for [A] ! knowing [C]=100)
let's find [FUGUR] using Inclusion–exclusion principle
\[\left| F \cup G \cup R \right|=\left| F \right|+\left| G \right|+\left| R \right|-\left| F \cap G \right|-\left| F \cap R \right|-\left| G \cap R \right|+\left| F \cap G \cap R \right|\]
with \[\left| F \right|=30, \left| R \right|=16,\left| G \right|= 19 \]
and \[\left| F \cap G \right|=12;\left| F \cap R \right|=11, \left| G \cap R \right|=6\]
finally \[\left| F \cap G \cap R \right|=4\]
now \[\left| F \cup G \cup R \right|=40\]
so (**)\[\left| A \right|=100-40\] !
It was so paainful :) :) ! If you need an explanation ! jusk ask :) !