<p>I got the number of possibilities for each set, but I have no idea what to do now.</p>
<p>Let one set be all positive integers that have three digits and a 1 in the ones place. Let set two be all positive integers that have three digits and a 2 in the tens place. How many are there in the union but not the intersection of the two sets?</p>
<p>I got 90 as the maximum possibility for each one, but I have no idea where to go from here.</p>
<p>The answers are
A. 162
B. 171
C. 180
D. 182
E. 191</p>
<p>Obviously we can eliminate C, D, E (90+90=180), but as for A and B? The answer key says that the answer is A, 162.</p>
<p>Let one set be all positive integers that have three digits and a 1 in the ones place. Let set two be all positive integers that have three digits and a 2 in the tens place. How many are there in the union but not the intersection of the two sets?</p>
<p>The numbers are of the form :xy1 and a2b
Possible numbers of form xy1 = 9<em>10</em>1 = 90
Possible numbers of form a2b = 9<em>1</em>10 = 90
Numbers present in both groups -->are of form m21 -->9 possible numbers
So number of numbers present in only union of the two sets but not in the intersection = 90 + 90 - 2*9 = 162.</p>
<p>We subtract twice of the intersection because:
A U B = A + B - ( A intersection B)
so A U B - (A intersection B) = A + B - 2*(A intersection B)</p>
<p>:D</p>
<p>wow. this is confusing me. i better go throiugh it again</p>
<p>Here's a simpler example, let set A be 1, 2, 3, 4, A, B, C and set B be 3, 4, 5, X, Y, Z. As you can see, set A has 7 elements and set B has 6 elements. 2 elements are common to both sets. The union of the two sets is 1, 2, 3, 4, A, B, C, 3, 4, 5, X, Y, Z. But since we don't allow duplicates, it's actually 1, 2, 3, 4, 5, A, B, C, X, Y, Z, or 11 elements. The intersection of the two is 3, 4 or two elements. The number of elements in the union but not the interesection is thus 11-2 or 9.</p>
<p>ohh alright. Thanks gatordude. But in the original problem, whats a fast way of counting all the elements in one set?</p>
<p>Possible numbers of form xy1 = 9<em>10</em>1 = 90
Possible numbers of form a2b = 9<em>1</em>10 = 90</p>
<p>xy1 numbers are 90 because x can be 1 to 9, y can be 0 to 9, and 1 is fixed.
a2b are 90 numbers because a can be 1 to 9, 2 is fixed, and b can be 0 to 9.
and you multiply them when you want both conditions to be satisfied.</p>
<p>spidey's explanation was perfect. Thanks =]</p>