Topic: Repeated and Non-Repeated Substitutions

Theorem 1. For the number of repeated substitutions

=

the formula is , here number of elements ,k -number of species.

1 - example. Two a. consists of one b and two c letters
plot all the duplicates for the tuple.
In this example, there are three (k = 3) and the number of letters is five (n = 5).
= 2 (two α). = 1 (one b ) and = 2 (two c ). The first two
interchange the positions of the letter (as well as the last two letters).
if we replace, new replacements will not be created. All repetitive
number of permutations

All thirty of these permutations are listed below:

aahcc, aachc, aacch, abacc. abcac, abcca ,
acubc.acacb, acbac, acbca, accab, accba ,
baacc, bacac, bacca, bcaac, beаса, bccaa .
eaabc, caacb, cabac, cabca, cacab, cacba ,
cbaac, cbaca, ebeaa, ccaab.ccaba, ccbaa

Non-repeated substitutions

Based on the concept of a set. the place in question here

we remind you that the substitutions do not repeat elements.
Therefore, such substitutions are unique (not repeated) positions can also be called substitutions.
All permutations for a given set of n elements
it is accepted to denote the number by .
For a set with one element {a}, there is only one position of the form a
it is obvious that there is a substitution:
Permutations from the elements of the two-element set {a,b}
using the permutation a for the set {a} with one element
we organize as follows: if element b is written after element a, then ab
has substitution, and if written before ba has substitution
we will be So, according to the multiplication rule, there are two permutations:
Permutations are arranged for the three-element set {a,b,c}
ab and ba are constructed for a two-element set {a,b}
substitutions can be used c of the given set
with three different ways of replacing the element ab and ba
can be placed: after their elements, elcments
between and before elements. Using the multiplication rule, three
six different positions for the set with elements {a ,b ,c}
we determine the formation of substitutions. They are:
abc. acb, cab, hac. bca, cba.
Given the four-element set {a,b,c,d}, the three-element set {a,b,c} d to each of the six substitutions made for the set
Note that there are four ways to place the element
then, according to the multiplication rule, we find Here are all the permutations:
abcd, abdc, adbc. dabc ,
acbd, acdb, adcb, dacb ,
cabd, cadb, cdab, dcab .
bacd, badc, bdac, dbac ,
bcad, hcda, bdca, dbca ,
cbad, chda, cdba, dcba.
we find Here are all the permutations:
Continuing in this way, “n is all space for a collection of elements all natural numbers with number of permutations from one to n It can be assumed that it is equal to the product of: . This the correctness of the hypothesis is proved in the following theorem 1.
Determining the product of the first n natural numbers in the form of n
accepted, that is, from the sign in such a sense
first K. Kramp published in 1808 on algebra
used in the manual.
When n = 1 in the expression , only the number 1 is involved,
therefore, 1!=1 is accepted as a definition. From this
except when n = 0 and n! the expression loses its meaning altogether.
However, 0!=1 is accepted as a definition.