<p>18.ABCDEFG
a list consists of all posiible three-letter arrangements formed by using the letters above such that the first leter is D and one of the remaining letter is A. if no letter is used more than once in an arrangement in the list and one three-letter arrangement is randomly selected form the list, what is the probability that the arrangement selected will be DCA?</p>
<p>ans ( 1 / 10 )</p>
<p>Having DA<em>(any other letter) is 5 possible combinations. Then having D</em>A is another 5 combinations. Therefore there’s a total of 10. DCA is just one possible combination out of 10. Hence, 1/10.</p>
<p>how many positive integers less than 1000 are multiples of 5 and are equal to 3 times an even integer ?</p>
<p>Okay, if you use logic here, there are only 100 integers less than 1000 that are multiples of 5, namely those integers that end in 5 or 0. Now, if you take 3 times any even integer, you’re going to get another even integer. Thus, you are left with only 50 integers, namely those ending in 0 (i.e. 10, 20, 30, etc.). Now, the only ones that are divisible by 3 are the ones whose digits add up to a number divisible by 3 (e.g. 120: 1+2+0=3). The only ones of the subset above that do that are 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360, 390, 420, 450, 480, 510, 540, 570, 600, 630, 660, 690, 720, 750, 780, 810, 840, 870, 900, 930, 960, 990. So, 33 numbers? I’m sure you see the pattern here.</p>