Permutations 排列

How many different ordered arrangements of the letters a, b, and c are possible? By direct enumeration we see that there are 6; namely, abc, acb, bac, bca, cab, and cba. Each arrangement is known as a permutation. Thus there are 6 possible permutations of a set of 3 objects. This result could also have been obtained from the basic principle, since the first object in the permutation can be any of the 3, the second object in the permutation can then be chosen from any of the remaining 2, and the third object in the permutation is then chosen from the remaining 1. Thus there are $3\times 2\times 1=6$ possible permutations.

Suppose now that we have n objects. Reasoning similar to that we have just used for the 3 letter shows that there are

$n\times (n-1)\times (n-2)\times ... \times 3\times 2\times 1= n!$

different permutations of the n objects.

A class in probability theory consists of 6 men and 4 women. An examination is given, and thestudents are ranked according to their performance. Assume that no two students obtain the same score.
(a) How many different rankings are possible? (b) If the men are ranked just among themselves and the women among themselves, how many different rankings are possible?
(a) As each ranking corresponds to a particular ordered arrangement of the 10 people, we see that the answer to this part is 10!=3,628,00.
(b) As there are 6! possible rankings of the men among themselves from the basic principle that there are $(6!)\times (4!)=(720)\times (24)=17,280$possible rankings in this case.          $\rule[0.02em]{1.0mm}{1.5mm}$