<p>This question was given as an example in the PSAT student guide for math.
How many different three-digit numbers greater than 240 can be formed by using three different digits from the set {1,2,3,4}?
(a) 24
(b) 16
(c) 14
(d) 12
(e) 10</p>
<p>I got the answer after using the long method and listing out each possibility; this took me 3 minutes. Is there a quicker way?</p>
<p>Well, it’s only a little quicker, but as you list, notice that there are only 3 choices for the hundreds place: 2, 3 or 4.</p>
<p>If you go with 2 ___ ___, you HAVE to put a 4 in the tens column. Then you have two options for the ones column.</p>
<p>If you go with 3 ___ ___, you have 3 choices for the tens place and two for the ones place…that makes 6 more combos.</p>
<p>Same thing if you go with 4 ___ ____ – 6 more combos.</p>
<p>Makes 14 all together…</p>
<p>^I believe that’s the quickest solution. You can also “complementary count” by counting all numbers beginning with 1 _ <em>, 21</em>, 23_ and subtracting that from 24. 24-10 = 14.</p>