<p>Straight from McGraw-Hill's book on SAT:</p>
<p>1) A 6-digit security code is made up of four 0s and two 1s. How many security codes of this type can be formed?</p>
<p>The answer is 15, but I don't know how to get to that number. The book's explanation was to just write out all the combinations and count them (ughhh).</p>
<p>2) There are 4 blue, 6 red, and 8 green marbles in a jar. What is the least number of marbles that can be removed without replacing, to guarantee that a chosen marble is red?</p>
<p>The answer is 13, but I thought it was 12. Why do I need to pick another marble to guarantee that I'll get a red marble? Are my chances of getting a red still not certain once I've picked out 4 blue and 8 green marbles?</p>
<p>I’ll start with the 2nd one: you have to actually draw the 13th marble! They are not asking “How many do you have to draw to be sure of getting it on your NEXT try?” – so you do have to go ahead and pick the 13th marble and then you are sure to get it.</p>
<p>As for the 1st one, here are two ways: </p>
<ol>
<li><p>Arranging items with duplicates: do 6!/(4!x2!) </p></li>
<li><p>But if you have never learned that much combinatorics, there is another way to think about it: Think of the problem as if you had 6 chairs (the digits) and two people to assign seats to (the ones). How many ways could you do it? You have 6 choices for the first person and then 5 for the next person. That makes 6 x 5 = 30 ways. But wait: the people are identical, so if you switch their places, it’s still the same arrangement…better divide the 30 by 2…</p></li>
</ol>
<p>In number 1, you are choosing 4 positions out of the 6. This is a combination. You can put 6 C 4 right into your calculator.</p>
<p>You can also think about it as choosing 2 positions out of the 6 (if you think of choosing the 1’s instead of the 0’s). So you can also do 6 C 2 in your calculator. You get the same answer.</p>
<p>If you are not familiar with combinations then an alternate way to get the result for problem 1 is to determine the number of unique ways to place the 2 1s into the the 6 slots.</p>
<p>It works like this:</p>
<p>(1) Place a 1 in the leftmost digit, and note that there are 5 possible ways to place the second 1.
(2) Place a 1 in the second leftmost digit, and note that there are 4 possible ways to place the second 1 (the leftmost digit is excluded because you’ve already counted that possibility above.</p>
<p>Continue with (3), (4) and (5).</p>
<p>You get 5 + 4 + 3 + 2 + 1 or 15.</p>