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## Here's the Solution to this Question

Let

A is set of students who are taking French

B is set of students who are taking business

C is set of students who are taking music

U is set of all students

Then we have:

$|U|=191, |A|=65, |B|=76, |C|=63$

$|A|\bigcap B|=36, |A|\bigcap C|=20, |B|\bigcap C|=18$

$|U|=|A|+|B|+|C|-|A\bigcap B|-|A\bigcap C|-|B\bigcap C|+|A\bigcap B\bigcap C|$

$|A\bigcap B\bigcap C|=191-65-76-63+36+20+18=61$

But $|A\bigcap B\bigcap C|\leq18$

So, let $|A\bigcap B\bigcap C|=10$

Then:

Who are taking French and music but not business:

$|A\bigcap C|-|A\bigcap B \bigcap C|=20-10=10$

Who are taking business and neither French nor music:

$|B|-|B|\bigcap C|-|A|\bigcap B|+|A\bigcap B \bigcap C|=76-18-36+10=22$

Who are taking French or business (or both):

$|A|+|B|-|A|\bigcap B|=95+76-36=135$

Who are taking music or french (or both) but not business:

$|A|+|C|-|A|\bigcap B|-|B|\bigcap C|+|A\bigcap B \bigcap C|=65+63-36-18+10=84$

Who are taking none of the three subjects:

$|U|-(|A|+|B|+|C|-|A\bigcap B|-|A\bigcap C|-|B\bigcap C|+|A\bigcap B\bigcap C|)=$

$=191-65-76-63+36+20+18-10=51$