what is a permutation?

huskyGdawg

New member
Joined
Mar 14, 2006
Messages
6
i missed a few days of school and i have no idea what a permutation is... its not in my math book... my teacher likes to go outside the book lately

ex: list all the permutations of 1,2, and 3
 
huskyGdawg said:
i missed a few days of school and i have no idea what a permutation is... its not in my math book... my teacher likes to go outside the book lately

ex: list all the permutations of 1,2, and 3

We designate the permutations of n things taken n at a time as nPn and the permutations of n things taken r at a time as nPr where P stands for permutations, n stands for the number of things involved, and r is less than n. To find the number of permutations of n dissimilar things taken n at a time, the formula is nPn = n! which is n factorial which means n(n-1)(n-2)(n-3).......3x2x1. Example: How many ways can you arrange the letters A & B. Clearly 2 which is 2 x 1 = 2. How many ways can you arrange the letters A, B & C in sets of three? Clearly 3P3 = 3 x 2 x 1 = 6. How many ways can you arrange A, B, C & D in sets of four? Clearly 4P4 = 4 x 3 x 2 x 1 = 24.

To find the number of permutations of n dissimilar things taken r at a time, the formula is nPr = n(n-1)(n-2)(n-3)..........(n-r+1). Example: How many 3-place numbers can be formed from the digits 1, 2, 3, 4, 5, and 6, with no repeating digit? Then we have 6P3 = 6x5x(6-3+1) = 6x5x4 = 120. How many 3-letter arrangements can be made from the entire 26 letter alphabet with no repeating letters? We now have 26P3 = 26x25x(26-3+1) = 26x25x24 = 15,600. Lastly, four persons enter a car in which there are six seats. In how many ways can they seat themselves? 6P4 = 6 x 5 x 4 x (6-4+1) = 6x5x4x3 = 360.

Another permutation scenario is one where you wish to find the permutations of n things, taken all at a time, when p things are of one kind, q things of another kind, r things of a third kind, and the rest are all different. Without getting into the derivation, nPn(p,q,r,s) = n!/(pxqxr). For example, how many different permutations are possible from the letters of the word committee taken all together? There are 9 letters of which 2 are m, 2 are t, 2 are e, and 1 c, 1 o, and 1 i. Therefore, the number of possible permutations of these 9 letters is 9P9(2,2,2,1,1,1) = 9!/(2x2x2x1x1x1) = 362,880/8 = 45,360.

In your case:

123
132
213
231
312
321
 
Top