<p>there are 5 different symbols:
A B $ D E
how many different arrangements, with $ never at either end?</p>
<p>Here is how you solve this:</p>
<p>First, figure how many different arrangements are possible, including the $ symbol at the ends.</p>
<p>We have 5 symbols to go in 5 places. </p>
<p>In the first space, we can put any of the 5 symbols. Now we have 4 symbols left. In the second space, we can put any of the 4 remaining. Then we have 3 left. We can put any of the three in the third space. Then we have 2 left. We can put either one in the 4th space. Then there's only one remaining symbol to go in the fifth space. </p>
<p>So we have a total of 5x4x3x2x1 = 5! = 120 different ways to arrange the symbols. </p>
<p>Now about the $ at the endpoints. How many of those arrangements have $ at either of the ends?</p>
<p>If $ is at the left end, there are 4 symbols left for the 2nd space, 3 for the 3rd, 2 for the 4th, and 1 for the 5th, so there are 4x3x2x1 = 4! = 24 ways to arrange the symbols with the $ at the left end.</p>
<p>Similarly, there are 24 ways to arrange the symbols with $ at the right end. </p>
<p>So the total number of arrangements WITHOUT $ at the end is given by:</p>
<p>TOTAL NUMBER WITHOUT $ AT AN END = TOTAL NUMBER OF ARRANGEMENT - NUMBER WITH $ AT AN END = 120 - 24 - 24 = 120 - 48 = 72.</p>
<p>There are 72 arrangements without $ at either end. </p>
<hr>
<p>About your graphing calculator, if you're using a TI-83 or something similar, this is what you do to use the combination and permutation functions. </p>
<p>If you want to know how many ways you can select a group of x items from a group of y items irrespective of the order of the items in the group of x items, you want to perform the function:</p>
<p>yCx or (y choose x) The bigger number goes first (number in total group)</p>
<p>For example, if you want to know how many ways you can select a group of three digits from 0,1,2,3,4,5,6,7,8,9 , you would enter 10C3. NOTE: since order doesn't matter in combinations, the groups (1,2,3), (2,1,3), (312), (1,3,2), (2,3,1), and (3,2,1) would only count for 1 combination, for example.</p>
<p>Let's say you have 10 friends and you can only bring 3 to dinner, but you love them equally so you have to pick a group of 3 at random. The number of DIFFERENT groups of 3 people you could bring out of that group of 10 would be 10C3. Let's say Alice, Bob, and Cathy are chosen. It doesn't matter if, for example, you pick them as Alice, Bob, and Cathy; Bob, Cathy, and Alice, etc. The order group of the three people in this case doesn't matter; they're all going to dinner. </p>
<p>The same wouldn't be true if you were picking 3 people from a group of 10 at random to be members of a council where one member is president, one vice-president, and one treasurer. Order in this case DOES matter, because the three people in the group are not equivalent. </p>
<p>In this case, we calculate 10P3, which give the total number of groups of 3 out of the total group of 10, and this takes order into account. For example, 10P3 would count (Alice, Bob, Cathy) and (Bob, Cathy, Alice) as two different groups. </p>
<p>As a rule, nPr is greater than or equal to nCr. </p>
<p>To compute nPr and nCr on your TI-83, enter your value for n in the work screen (total number of elements in your larger group), hit MATH, use > to scroll over to PRB, then nPr and nCr are 2nd and 3rd in the list. Select the one you want with ENTER, then you'll be back in the work screen, then enter your value of r (number of elements in the smaller group you're selecting from the bigger group). Then press enter and there's your answer!</p>
<p>If you don't have these operations on your calculator, you can use the following identities:</p>
<p>nPr = n!/(n-r)!</p>
<p>nCr = nPr/rPr = [n!/(n-r)!]x[1/r!] = n!/[r!(n-r)!]</p>
<p>Where a! = a x (a-1) x (a-2) x .... 3 x 2 x 1 (I'm assuming a > 3 here just to show how to calculate this)</p>