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Best answer
7 votes
7 votes
$  n(A\cup B) = n(A) + n(B)  - n(A \cap B)$

$\qquad  =  \lfloor \frac{1000}{7} \rfloor + \lfloor \frac{1000}{11} \rfloor -  \lfloor \frac{1000}{7*11}\rfloor = 142+90-12 =  220$

$n((A - B) \cup (B-A)) = n(A\cup B) - n(A \cap B)$

$\qquad  =220 -12 = 208$
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3 votes
3 votes
$n\left ( either A or B is true \right )=n(A \bigcup B)=n(A)+n(B)-n(A\bigcap B)$

$n( exactlyone of A or B is true)$ = $n((A-B)\bigcup (B-A))=n(A)+n(B)-2*n(A\bigcap B).$

So,

n(divisible by exactly one of 7 or 11)=142+90-2*12= 232-24=208
Answer:

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