Sat Q

Test Preparation
1

<p>5 cards are placed in a row so that one card is never at either end. How many different arrangements are possible?</p>

<p>my work:
3<em>4</em>3<em>2</em>1=72
The answer is 72.</p>

<p>Is there any other way to solve this?</p>

2

<p>Could it be "72!" (which is basically the same thing you did)?</p>

3

<p>18(4)=72
5 cards minus 1 card is four cards. If the 1 card cannot be at either end, then it has 3 possible positions in the row. At each position there are six arrangements of the other four cards. Finally, you multiply the 6 x 3 x 4 to get 72. </p>

<p>Actually, I have no idea...</p>

4

<p>5!-4!*2 = 72
5! : arrangements of five cards without the condition.
If "the card" IS at the front, 4! is possible. (another 4! with same thinking in the back.)
(# satisfying condition) = (whole #) - (# that doesn't satisfy condition)
Just like "Male = Human - Non-male"</p>

<p>so x = 5! - 4! - 4! = 5! - 4!*2 = 72.</p>