Theorem 1.
A space for a set with n elements
the number of permutations is equal to n , that is, P = n .
Proof. From the method of mathematical induction to prove the theorem
we use The correctness of the basis, that is, the confirmation of the theorem for n = 1
we saw the correctness above. Proof of the theorem for induction transition
we assume that it is true for some natural n = k, that is, = k
let it be Clearly, a set with (k + 1) elements into a set with k elements
a new (k + 1)- element can be created by inserting. This
(k + 1)- element for a set with k elements for all k! place
insert into each of the permutations in (k +1) different ways
can:
1 - before the element,
between elements 1 and 2,
between elements 2 and 3,
between (k — 1) - and k - elements,
k - after the element.
So, by the multiplication rule, it has (k + 1) elements
total for the set k!(k + 1) = (k + 1)! create substitutions
will be, that is, = (k + 1)!
Example 1.
Take five places out of five spectatorsfind the number of possibilities (options).If we mark the audience with letters a, b, c, d, e, thenWe get a set of viewers T — {a,b,c,d,e}. To each of the possibilities (options) of placing the audience in seatsThe audience is some replacement of the elements of the set Tfits. Since the set T has five elements, by Theorem 1basically, = 1 * 2 * 3 * 4 * 5 = 120. So, the number of possibilities for five viewers to occupy five places is equal to 120.
Conclusion
We have tried to cover the topic thoroughly by editing definitions and various examples on the subject of repeated and non-repeated permutations.
thank you for your attention
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