<p>I desperately need help with those concepts such as the difference between them and how to tackle those problems. I'm pretty weak in this area so any help would be appreciated:). BUT the March Sat is tomorrow so I'm hoping for a response tonight. Thanks!</p>
<p>Okay suppose you have n objects and need to choose r.</p>
<p>Let’s use the example “abcd” and you need to choose two letters,</p>
<p>Permutations: Order does NOT matter. (meaning ab and ba are separate)</p>
<p>If repetition is allowed (meaning aa is allowed) the formula is n^r *Ex. 4^2 = 16 *
If repetition is not allowed (meaning aa is NOT allowed) the formula is n<em>r *Ex. 4</em>2 = 8 *</p>
<p>Combinations: Order does matter (meaning ab and ba are the same)</p>
<p>If repetition is allowed, don’t worry about it. Too difficult for the SAT.
If repetition is not allowed the formula is n!/((n-r)!r!) *Ex. 4!/(4-2)!(2!) = 6 *</p>
<p>Meadow 36 you made that way too complicated. let’s say you have ABCDE and you need to find how many different combinations of those letters there are without repeating them. The letter in the first place can be any of the 5 letters, for the second spot, it can be any of the 4 letters that remain (remember one is already in the first spot) and for the 3rd spot there are three letters to choose from and so on. To find the number of combinations you multiply 5 x 4 x 3 x 2 x 1=120</p>
<p>In this scenario lets imagine that you have the letters ABCDE and you are minding how many combinations there are but the letters can be repeated such that you can have AACDE or BBBBB. for the first slot, there are 5 letters to choose from, since the second spot can be the same letter as the first slot, it also has 5 letters that it can be, and continues for however many things there are. 5 x 5 x 5 x 5 x 5 = 3125</p>
<p>Finally lets say there are 10 students in a class and 2 are randomly selected to win a prize. How many DIFFERENT combinations of students can there be that win the prize. well the first winner can be any of the 10 students, and the second winner is any of the remainding 9 students. so 10 x 9= 90. But wait, if student A wins the first prize and student B wins the second prize it is the same as if student B one the first prize and A the second. (AB is the same as BA) so we must divide our answer by two. 90/2=45 combinations of winners</p>
<p>both wonderful explanations ^</p>
<p>In permutation order matters and in combinations order doesn’t matter… (to meadow)</p>
<p>Hmm yea Josh66 gave a very nice intuitive explanation. And yea, sanateria is right -_- wow.</p>
<p>Understanding Josh’s explanation is better than memorizing what I gave lolz</p>
<p>Hey guys,</p>
<p>Can you explain why do we need to multiply for possibilities and not add? I don’t understand lol. I understand that after the 1st try there is one less in the second and such but why multiply and not add? Please help!</p>
<p>Thanks!</p>