<p>Jeffrey,</p>
<p>A A _ _ 7<em>6 = 42 ways (The last two letters shouldn’t be the same in your counting; otherwise AABB gets counted twice)
A _ A _ 7 7=49 ways
A _ _ A 7</em>6=42 ways (The middle two letters shouldn’t be the same; otherwise ABBA gets counted twice)
_ A A _ 7<em>7=49 ways
_ A _ A 7</em>7=49 ways
_ _ A A 76=42 ways (The first two letters shouldn’t be the same; otherwise BBAA gets counted twice)
A A A _ 7 ways
A A _ A 7 ways
A _ A A 7 ways
_ A A A 7 ways
A A A A 1 way</p>
<p>(49<em>3) + (42</em>3) + (7*3) + 1 = 302 ways for alphabet A to meet the condition.</p>
<p>302 * 8 = 2416 ways</p>
<p>Yes, haha. I did figure that out. It was really irritating at first because the answers didn’t match. Double-counting only occurs for even-letter-codes greater than 4, right? (4, 6, 8, 10,…)</p>
<p>I absolutely love your questions, pckeller. :)</p>
<p>In a primary election, there are four candidates for mayor, five candidates for city treasurer,
and two candidates for county attorney. In how many ways may voters mark their ballots
a) if they vote in all three of the races?
b) if they exercise their right not to vote in any or all of the races?</p>
<p>Correct for both a) and b), Jeffrey. :)</p>
<p>a) 4 x 5 x 2 = 40
b) 5 x 6 x 3 = 90 (Think of “omit” as an extra choice for each race)</p>