<p>hi. i can't seem to get how to solve this math problem. can someone help me?</p>
<p>if 5 cards (numbered 1,2,3,4,5) are placed in a row so that card number 3 is never at either end, how many different arrangements are possible?</p>
<p>The left-most card can be 1/2/4/5 --> four possibilities
The right-most card can be 1/2/4/5 BUT the first card has already been chosen --> three possibilities
The three center cards can be any of the three remaining cards --> three, two, one possibilities, respectively</p>
<p>right x left x center1 x center2 x center3
4 x 3 x 3 x 2 x 1 = 72</p>
<p>oh okay. thanks!</p>