<p>If anyone could explain how to approach this problem, I would greatly appreciate it</p>
<p>Maybe I should have posted the question, lol. </p>
<p>(I am changing the wording a bit) If there are five different playing cards and one playing card cannot be at either end of the arrangement (assume a 1-2-3-4-5 arrangement), how many arrangements are possible?</p>
<p>Helpppppppppppppppppp</p>
<p>This is how I did it, </p>
<p>A B C D E (Let C represent the gray card)</p>
<p>... So you can do A C B D E , A C B E D , A C E B D, A C E D B, A C D E B, A C D B E.</p>
<p>Since writing out all the combination would be very, very time consuming, try to see a pattern with this first set. With two controls, A in the Front and C (gray) second, you can have 6 different possibilities. If you don't see the pattern, you can do another set. This time with A in the front and C in the middle. you have A B C D E, A B C E D, A E C B D, A E C D B, A D C B E, A D C E B. So again you have 6 possibilities. So now you see that maintaining A as a control will give 6 possibilities. Well how many times does this happen? You'll get 6 combination with A front and C second, 6 more with A front, C third, and 6 more with A front and C second to last (remember C can't be last or first). So with A going first we get a total of 24 combination... Now just multiply that by 3 (B being in the front, D being in the front, and E being in the front). </p>
<p>I'm sure there is a "more" mathematical way, but this is just how I did it.</p>
<p>Actually, nvm. I get it now. :P I suck at these.</p>
<p>No its 72.</p>
<p>ahhhh....thanks!! Is there a way to explain this mathematically, however?</p>
<p>????????????????????</p>
<p>Given 5 things, you would normally have 5! = 120 arrangements ... here, though, one card must be in slot 2, 3, or 4. Each choice of 2, 3, or 4 leaves 4! arrangements for the other cards, so the grand total is 4! + 4! + 4! = 72.</p>
<p>If they didn't give you that strange condition about a card not being allowed at either end, the problem is simple. I am assuming you know that the solution would then simply be 5 x 4 x 3 x 2 x 1.</p>
<p>Now we will incorporate this given condition. Let's assume that the first card we place in one of the five "spots" is going to be the one that cannot be at either end. So, we have to place this card in spots 2,3 or 4 (not 1 or 5 because they are the ends). </p>
<p>So there are 3 possible areas for this first card -> 3</p>
<p>The next card has 4 possible areas, since the first one already took up a spot -> 4</p>
<p>The next card -> 3</p>
<p>The next card -> 2</p>
<p>The next card -> 1</p>
<p>So the solution is: 3 x 4 x 3 x 2 x 1 = 72</p>
<p>^^^^^ Correctomundo. :)</p>
<p>Ahhh, thanks guys, i get it now!</p>