Let's assume that are given N elements (numbers, objects, etc.). It is required to learn in how many ways these N elements it is possible to row. In more exact terms, it is required to calculate quantity of possible combinations from these elements.
Instruction
1. If it is supposed that in a row enter all N elements, and any does not repeat, then it is a task about the number of shifts. A solution can be found a simple reasoning. On the first place among any of N elements can be, therefore, N options turn out. On the second place — any, besides, which was already used for the first place. Therefore, for each of N already found options is (N - 1) options of the second place, and the total number of combinations becomes N* (N - 1). The same reasoning can be repeated for other elements of a row. For the latest place there is only one option — the last remained element. For penultimate — two options, and so on. Therefore, for a row from N of not repeating elements the number of possible shifts is equal to the work of all integers from 1 to N. This work is called a factorial of number N and N is designated! (it is read "en a factorial").
2. In the previous case the quantity of possible elements and the number of places of a row coincided, and their number was equal to N. But the situation when it among less places, than is available possible elements is possible. In other words, quantity of elements in sample to equally some number M, and M <N. In this case the problem of determination of quantity of possible combinations can have two various options. First, can be required to count total number of possible ways which it is possible to build in a row the M elements from N. Such ways are called placements. Secondly, the researcher can be interested in number of ways as which it is possible to choose the M elements from N. At the same time the order of arrangement of elements is not important any more, but any two options have to differ among themselves at least with one element. Such ways are called combinations.
3. To find the number of placements on M elements from N, it is possible to resort to the same way of reasonings, as well as in a case with shifts. On the first place here N elements, on the second (N - 1) can still be and so on. But for the last place the quantity of possible options equals not to unit, and (N - M + 1) as when placement is finished, will remain still (N - M) unused elements. Thus, the number of placements on M elements from N equals to the work of all integers from (N - M + 1) to N, or that the same, to private N! / (N - M)!.
4. It is obvious that the quantity of combinations on M elements from N will be less number of placements. For each possible combination there are M! the possible placements depending on an order of elements of this combination. Therefore, to find this quantity, it is necessary to divide number of placements on M elements from N into N!. In other words, the quantity of combinations on M elements from N is equal to N! / (M! * (N - M)!).