<p>The official SAT guide first edition.
Pratice Test 2, Section6, Problem 18. (page 476 problem 18)</p>
<p>CRUDE interpretation of the problem.
If 5 cards:4 Spades and 1 Ace are arranged in a row in such a way that Ace is never at either end, how many different arrangements are possible?</p>
<p>I reckon this problem is based on Permutations.
I tried solving it in this way: Wrong answer
5! = 120 (gives total number of arrangements possible)
120 - 2 = 118 (since the card must never be in either ends)</p>
<p>Other method: Right answer
4P3 = 60 (since others have 5 options for 3 slots)
4P2 = 12 (since ends have 4 options for 2 slots</p>
<p>60+12 = 72 = CB answer.</p>
<p>Help!!! why ain't I getting answer in the first method?</p>
<p>cause you have to do the ace and the other cards spearately or else its just like putting each card in any place. subtractign two has no significance</p>
<p>Basically there are three sets of arrangements. That is the grey box can be in the middle of to either of the spaces the right or left. L
Let’s take the first situation. If the card is in the middle, there are 4 possibilities for the first box, three for the second, two for the next, and one for the last square.
There are three possible arrangements of this, so the answer is:</p>
<p>Basically if you subtract 2 you’re saying that the position of the other 4 cards does not matter. But it doesn’t work that way because you counted all possible permutations with 5! then all cards count separately. Hence, what you need to do is 5! minus 2 (4!) (because the ace cards will be in either ends and the other 4 cards will be put separately in other positions). so 5! - 2(4!) = 120 - 48 = 72.</p>
<p>Here’s a simpler way that doesn’t even involve factorials. Imagine five spaces for the cards</p>
<hr>
<p>Because an ace can’t be on the far left end, we can put any one of 4 spades there. An ace also cannot be on the far right, but we already used a spade, so one of 3 spades can go there. </p>
<p>Now onto the middle three spaces. Any one of the 3 cards left (1 ace, 2 spades), can go in a given middle space, leaving 2 possibilities for the next middle space, and 1 possibility for the last middle space. Now just multiply all those possibilities together</p>