<p>This is one page 476 in the Collegeboard blue book...</p>
<p> </p>
<p>If the five cards shown above are placed in a row so that the card that is presently in the middle is never at either end, how many different arrangements are possible. </p>
<p>The answer is 72....but i don't know how to get it?</p>
<p>its um 4 x 3 x 2 x 1 x 3 its too hard to explain but ill try, </p>
<p>? ? ? ??</p>
<p>the first card has to be picked from a total of 4 since the black one cant be in the front </p>
<p>so its </p>
<p>4 ? ? ? ?</p>
<p>then the last one there are 3 cards to choose from since the black one cannot be on the end so then tis </p>
<p>4 ? ? ? 3</p>
<p>then you have three cards left including the black and the other two so
just plug them in</p>
<p>4 3 2 1 3</p>
<p>BOOm then multiply
and its 72 its kinda confusing but thats how u do it
i think</p>
<p>Ok Cool Thanx</p>
<p>Okay, this is a translation of what he is doing.
4!x3
Why 4!: For a moment, pretend that the middle one does noe exist. Okay. You got 4 cards that can go in x arrangements. So, how do you do a problem normally like that? 4! (4x3x2x1) and you get 24. The 'killer rabbit' (<em>shock</em> I'm transforming into Adam Robinson! lol) is that 24 is the answer, because it may 'seem'' the middle one is ComPLETLY not counted, but hence it is:</p>
<p>Why x3: Okay, let's draw the problem again
OOOOO okay. Reconfigue it to show that the middle O CANNOT go into the 2 sides: XOOOX So, now you have to find the middle value, which is 4!x3 which happens to be 72.</p>
<p>Now, this was done analyzing the questions for over 3 mintes. If I did this in a test, I would have got it wrong also.</p>