A more general statement of the rule of product

If an object a₁ can be chosen in m₁ ways,
an object a₂ can be chosen in m₂ ways,

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an object a_k can be chosen in m_k ways,

and if the choice of an object a₁ does not affect the number of choices of a₂, a₃,…a_k, the choice of a₂ does not affect the number of choices of a₃, a₄,….a_k, and so on then the choice of the ordered set of objects (a₁, a₂,…a_k) in that order can be accomplished in

m₁ * m₂ * m₃ * ….. * m_k ways.

Example:
Signs are made which consist of a geometrical figure (a circle, square, triangle, hexagon [‘heksэgэn] – шестиугольник), a Russian letter and a decimal digit. How many signs can be mede? Such as

B6 K8

At first step we can choose a geometric figure. This can be done in 4 ways since we have a total of 4 figures. Then we can choose one of 32 Russian letters and finally one of 10 digits.

This brings the total to 4 x 33 x 10 = 1 320 combinations.

Suptopic: Permutations without repetitions:

The general statement of the problem is as follows: There are n-digit objects M = {a₁, a₂,…a_n}. Usually these objects are considered to be just numbers.
How many k – arrangements (ordered k – subsets) can be made out of these elements?

Two arrangements are considered distinct if they differ even by a single element or have a different sequence of elements.

Example: M = {a, b, c}.

2 – arrangement of the set M:
(a, b), (b, a), (a, c), (c, a), (b, c), (c, b).

3-arrangements of the same set:

(a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a).

Such arrangements are termed permutations without repetitions.

The product 1 * 2 * 3 * …..* n is denoted by n! (is read: n – factorial).
It is agreed to consider 0! = 1.

The number of k-permutations out of n distinct objects is denoted by A_n^k.

The first element of any k – permutation can be chosen in n ways, the second element – in n – 1 ways and so on. Factually its k-th element can be chosen in n – k + 1 ways.

Applying the rule of product the necessary number of times we obtain:

A_n^k = n * ((n – 1) * (n – 2) * …. * (n – k + 1) = (1)

This is number of k – arrangements of n distinct elements.

When n = k then we obtain out of expression (1):

A_nⁿ = P_n = = = = n! (2)

Let us write down an important relation for the number of k – permutations out of n elements;

= + k * ; (3)

Relations of this type are called recursive (recurrence) relations.

We can prove that this equality is true in such a way:
If element 1 does not belong to a k – arrangement X, then X is a k-arrangement of the (n – 1)- set which is M – {1} (that is of the set M without the element 1).