Math Help

Test Preparation SAT Preparation
1

<p>Ok So I'm shaky on permutations so I would greatly appreciate the help.</p>

<p>Here's a question I made (based on a real one)</p>

<p>There are 5 different colored cards
Black
Blue
Red
Green
Yellow</p>

<p>How many combinations are possible between these 5 cards to put them in 5 slots. The only requirement is that the Black card be in the 1st or last slot.</p>

<p>can someone go through this step by step (and show how to do it using the permutation/combination forumla?)</p>

<p>Thanks in advance.</p>

2

<p>Imagine 5 slots</p>

<hr>

<p>2 scenarios Black card is in the first slot or black card is in the last:</p>

<p>Black Card in the first slot looks like this</p>

<p>1x4x3x2x1 </p>

<p>Black Card last slot looks like this</p>

<p>4x3x2x1x1 .</p>

<p>Add them together and voila!</p>

3

<p>^wait buy why does a 1 fill in the slots?</p>

4

<p>and also, how do u do this using the formula?</p>

5

<p>5!-(4!x2) its in kev's file i mentioned, its from the bb, test 2</p>

6

<p>I edited my post to have it make more sense. Basically, each number represents the possible number of cards which can be put in each slot.</p>

<p>PS: formulas are for sissies</p>

7

<p>@orange - what test, section, and number is it?</p>

8

<p>oh and the answer orange has is different from urs</p>

9

<p>no its the same concept. he just determined the total number of balck combinations
i did the total - black</p>

<h1>18 on test 2 section ?</h1>

10

<p>i'm doing the total too and u have different answers?</p>

11

<p>so what's the correct answer?</p>

12

<p>48
10chara</p>

13

<p>@pawn
what i dont get is why the first or last slot isn't 2 since there are TWO possibilities: 2 black cards
edit: wait theres 1 black card</p>

14

<p>There are 2 slots possible for the Black card,4 for the Blue one,3 for the red and 2 for the green one
,and 1 for the yellow one .
2x4x3x2x1 = 48 possible combinations.</p>

15

<p>That's an interesting way of looking at it.</p>

16

<p>U cud just ignore the black card for the time being. So ,that leaves u with 4 cards.Thus, the possible no. of combinations with these 4 cards will be 4x3x2x1=24
Now the black card cud either be at first or at last of each of the above combinations. So, there will be 24 combinations with black at first and 24 other combinations with black at last position.
Hence, total no. of combinations=24+24=48</p>