<p>From CB's official study guide book, #18, p. 476:</p>
<p>(They have 5 squares with different designs in them ... I think it really doesn't matter what is being used as the symbols)</p>
<p>A B C D E</p>
<p>If the 5 cards shown above are placed in a row so that C is never at either end, how many different arrangements are possible?</p>
<p>Answer: 72</p>
<p>I can get 72 by doing 3<em>2</em>4<em>1</em>3, but 1) I'm not sure if that's applicable in this scenario and 2) I'm not confident on the logic used in this one since you only have those 5 cards to use. Any help would be greatly appreciated.</p>
<p>ok... what's wrong with that logic...</p>
<p>4 possible cards... on the first end... excluding c... other end can have 3 possible choices... because it can't be c or the card used on the other side... one middle card must be 1 and in the other cases u have 3 choices just not c or the card used on the left side, and 2 not c or the card used on the other side, i believe someone asked this question before</p>
<p>ok, so think of the five positions as five slots you can put any cube into. Like this:</p>
<hr>
<p>to get the total # of possiblities, you multiply the # of possiblities for ea. slot. if C can't be in either end, when you place down the first cube in the first slot you have four options: A, B, D, or E.
4 _ _ _ _</p>
<p>for the last slot, you have three options, anything but C and the block you put down for the first slot:</p>
<p>4 _ _ _ 3</p>
<p>for the second slot you have three options: C and the other two blocks you didn't use to fill the first and last slot.</p>
<p>4*3 _ _ 3</p>
<p>for the third, there's still two blocks you haven't used, leaving only one possibility for the fourth</p>
<p>4<em>3</em>2<em>1</em>3= 72</p>
<p>voila! sorry if that was confusing. oh i just realized it's cards and not cubes/blocks. ah well too lazy to go back and fix it. :D</p>
<p>Thanks --- these explanations helped! :)</p>