How many primes are less than or equal to 100 by Inclusion Exclusion Rule And by using Mathematica

Let P1=Integers that are less than or equal 100 and divisible by 2

P2=Integers that are less than or equal 100 and divisible by 3

P3=Integers that are less than or equal 100 and divisible by 5

P4=Integers that are less than or equal 100 and divisible by 7

Thus number of primes less than or equal to 100= 4+N(P'1 P'2 P'3 P'4)

N(P'1 P'2 P'3 P'4)= N-N(P1)-N(P2)-N(P3)-N(P4)-N(P1P2)-N(P1P3)-N(P1P4)-N(P2P3)-N(P2P4)-N(P3P4)-N(P1P2P3)-N(P1P2P4)-N(P2P3P4)-N(P1P3P4)-N(P1P2P3P4)

$99-\left[\frac{100}{2}\right]-\left[\frac{100}{3}\right]-\left[\frac{100}{5}\right]-\left[\frac{100}{7}\right]+\left[\frac{100}{2·3}\right]+\left[\frac{100}{2·5}\right]+\left[\frac{100}{2·7}\right]+\left[\frac{100}{3·5}\right]+\left[\frac{100}{3·7}\right]+\left[\frac{100}{5·7}\right]$
$-\left[\frac{100}{2·3·5}\right]-\left[\frac{100}{2·3·7}\right]-\left[\frac{100}{3·5·7}\right]-\left[\frac{100}{2·5·7}\right]+\left[\frac{100}{2·3·5·7}\right]$

*(We put 99 because number [1] does not count as a prime number)*

 Of course we should take the floor for these fractions...

Now=99-50-33-20-14+16+10+7+6+4+2-3-2-1-0+0

=21

So , number of primes less than or equal to 100=> 21+4=25

Now by using mathematica...