Method 1: Conventional Approach

$1260=2^2\times 3^2\times 5\times 7$

We need to find the number of positive integers 1260 which are not divisible by 2, 3, 5 or 7.

Let A be the set consisting of multiples of $2\le 1260$;
B be the set consisting of multiples of 3;
C be the set consisting of multiples of 5 & D of multiples of 7.

$A=\frac{1260}{2}=630,$
$B=\frac{1260}{3}=420,$
$C=\frac{1260}{5}=252,$
$D=\frac{1260}{7}=180$
$A\Pi B=\frac{1260}{2\times 3}=210A\Pi C=\frac{1260}{2\times 5}=126A\Pi D=\frac{1260}{2\times 7}=90$

$B\Pi C=\frac{1260}{3\times 5}=84$
$B\Pi D=\frac{1260}{3\times 7}=60C\Pi D=\frac{1260}{5\times 7}=36$
$A\Pi B\Pi C=\frac{1260}{2\times 3\times 5}=42$
$A\Pi B\Pi D=\frac{1260}{2\times 3\times 7}=30$
$A\Pi C\Pi D=\frac{1260}{2\times 5\times 7}=18$
$B\Pi C\Pi D=\frac{1260}{3\times 5\times 7}=12$
$A\Pi B\Pi C\Pi D=\frac{1260}{2\times 3\times 5\times 7}=6$

So, $A\cup B\cup C\cup D=630+420+252+180-210-126-90-84-60-36+42+30+18+12-6=972$

So, 1260 - 972 = 288 integers are relatively prime to 1260.

Method 2: Shortcut: Using the concept of Euler's number

The number of positive integers $\le 1260$ which are relatively prime to 1260 can be directly found by finding the Euler's number of 1260
$=N(1-\frac{1}{2})\times (1-\frac{1}{3})\times (1-\frac{1}{5})\times (1-\frac{1}{7})$
$=1260\times \frac{1}{2}\times \frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}=288$

(where 2,3,5,7 are the prime factors of 1260)