Very nice post! +1.
pepemelo wrote:
Hi,
You should use a permutation when the order in which you are suppose to choose a number of objects from a set matters. As an example: In how many ways is it possible to arrange the letters of the word CAT in different 2-letter groups, where CA is different than AC (i.e., the order matters)?
\(3P52=\frac{3!}{(3-2)!}=\frac{3.2.1}{1!}=6\). The general permutation formula is given by \(nPm=\frac{n!}{(n-m)!}\). When both elements of the permutation are equal), \(nPn=n!\).
If the order in which the objects are chosen doesn't matter, you use combinations - the formula is very similar to the permutations formula, but you find one more factorial in the denominator: \(nCm=\frac{n!}{(n-m)!m!}\). Taking the same example, how many different 2-letter groups is it possible to get from the word CAT, considering that CA is the same as AC (i.e., the order doesn't matter)?
\(3C2=\frac{3!}{(3-2)!2!}=\frac{3.2.1}{1!2.1}=\frac{6}{2}=3\)
If there is repetition, i.e, if the is more than one particular element in the set, you should divide the permutation/combination value by the factorial of the number of objects that are identical. Examples:
How many different 5-letter words can be formed from the word APPLE? (note you have 2 Ps).
\(\frac{5P5}{2!}=\frac{5.4.3.2.1}{2.1}=\frac{120}{2}=60\)
How many different 6-digit numbers can be written using all of the following six digits: 4,4,5,5,5,7? (here you have 2 fours and 3 fives).
\(\frac{6P6}{2!3!}=\frac{6.5.4.3.2.1}{2.1.3.2.1}=\frac{720}{12}=60\).
Hope this helps!