PERMUTATIONS WITHOUT REPETITION: FORMULAS, PROOF, EXERCISES, EXAMPLES - DUDAS - 2025

A permutation without repetition of n elements is the different groups of different elements that can be obtained from not repeating any element, only varying the order of placement of the elements.

To find out the number of permutations without repetition, the following formula is used:

Pn = n!

Which expanded would be Pn = n! = n (n - 1) (n - 2)… (2) (1).

So in the previous practical example it would be applied as follows:

P4 = 4 * 3 * 2 * 1 = 24 different 4-digit numbers.

These being the 24 arrays in total: 2468, 2486, 2648, 2684, 2846, 2864, 4268, 4286, 4628, 4682, 4826, 4862, 6248, 6284, 6428, 6482, 6824, 6842, 8246, 8264, 8426, 8462, 8624, 8642.

As can be seen, there is no repetition in any case, being 24 different numbers.

Demo and formulas

24 Arrangements of 4 different figures

We are going to analyze more specifically the example of the 24 different 4-digit arrangements that can be formed with the digits of the number 2468. The number of arrangements (24) can be known as follows:

You have 4 options to select the first digit, that leaves 3 options to select the second. Two digits have already been set and 2 options remain for selecting the third digit. The last digit has only one selection option.

Therefore, the number of permutations, denoted by P4, is obtained by the product of the selection options in each position:

P4 = 4 * 3 * 2 * 1 = 24 different 4-digit numbers

In general, the number of different permutations or arrangements that can be performed with all the n elements of a given set is:

Pn = n! = n (n - 1) (n - 2)… (2) (1)

The expression n! it is known as n factorial and means the product of all the natural numbers that lie between the number n and the number one, including both.

12 Arrangements of 2 different figures

Now suppose you want to know the number of permutations or two-digit numbers that can be formed with the digits of the number 2468.

These would be 12 arrangements in total: 24, 26, 28, 42, 46, 48, 62, 64, 68, 82, 84, 86

You have 4 options to select the first digit, that leaves 3 digits to select the second. Therefore, the number of permutations of the 4 digits taken two by two, denoted by 4P2, is obtained by the product of the selection options in each position:

4P2 = 4 * 3 = 12 different 2-digit numbers

In general, the number of different permutations or arrangements that can be performed with r elements of the n in total in a given set is:

nPr = n (n - 1) (n - 2)…

The above expression is truncated before playing n !. To complete n! from it we should write:

n! = n (n - 1) (n - 2)… (n - r)… (2) (1)

The factors that we add, in turn, represent a factorial:

(n - r)… (2) (1) = (n - r)!

Thus, n! = n (n - 1) (n - 2)… (n - r)… (2) (1) = n (n - 1) (n - 2)… (n - r)!

From here

n! / (n - r)! = n (n - 1) (n - 2)… = nPr

Examples

Example 1

How many different 5-letter combinations of letters can be constructed with the letters of the word KEY?

We want to find the number of different letter combinations of 5 letters that can be built with the 5 letters of the word KEY; that is, the number of 5-letter arrays involving all the letters available in the word KEY.

N ° of 5 letter words = P5 = 5! = 5 * 4 * 3 * 2 * 1 = 120 different 5-letter letter combinations.

These would be: CLAVE, VELAC, LCAEV, VLEAC, ECVLAC… up to 120 different letter combinations in total.

Example 2

You have 15 numbered balls and you want to know How many different groups of 3 balls can be built with the 15 numbered balls?

You want to find the number of groups of 3 balls that can be made with the 15 numbered balls.

N ° of groups of 3 balls = 15P3 = 15! / (15 - 3)!

N ° of groups of 3 balls = 15 * 14 * 13 = 2730 groups of 3 balls

Solved exercises

Exercise 1

A fruit store has an exhibition stand consisting of a row of compartments located in the entrance hall to the premises. In one day, the greengrocer acquires for sale: oranges, bananas, pineapples, pears and apples.

a) How many different ways do you have to order the exhibition stand?

b) How many different ways do you have to order the stand if, in addition to the fruits mentioned (5), you received on that day: mangoes, peaches, strawberries and grapes (4)?

a) We want to find the number of different ways to order all the fruits in the display row; that is, the number of arrangements of 5 fruit items that involve all the fruits available for sale on that day.

N ° of stand arrangements = P5 = 5! = 5 * 4 * 3 * 2 * 1

N ° of stand arrangements = 120 ways to present the stand

b) We want to find the number of different ways to order all the fruits in the display row if 4 additional items were added; that is, the number of arrangements of 9 fruit items that involve all the fruits available for sale on that day.

N ° of stand arrangements = P9 = 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

N ° of stand arrangements = 362,880 ways to present the stand

Exercise 2

A small food outlet has a plot of land with enough space to park 6 vehicles.

a) How many different ways of ordering the vehicles in the plot of land can be selected?

b) Suppose that a contiguous plot of land is acquired whose dimensions allow 10 vehicles to be parked. How many different forms of vehicle arrangement can be selected now?

a) We want to find the number of different ways of ordering the 6 vehicles that can be housed in the plot of land.

No. of arrangements of the 6 vehicles = P6 = 6! = 6 * 5 * 4 * 3 * 2 * 1

Number of arrangements of the 6 vehicles = 720 different ways of ordering the 6 vehicles in the plot of land.

b) We want to find the number of different ways of ordering the 10 vehicles that can be housed in the plot of land after the expansion of the plot of land.

N ° of arrangements of the 10 vehicles = P10 = 10!

No. of vehicle arrangements = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Number of arrangements of the 10 vehicles = 3,628,800 different ways of ordering the 10 vehicles in the plot of land.

Exercise 3

A florist has flowers of 6 different colors to make floral flags of nations that have only 3 colors. If it is known that the order of colors is important in the flags, a) How many different flags of 3 colors can be made with the 6 available colors?

b) The seller buys flowers of 2 additional colors to the 6 he already had, now how many different flags of 3 colors can be made?

c) Since you have 8 colors, you decide to expand your range of flags. How many different 4-color flags can you make?

d) How many of 2 colors?

a) We want to find the number of different flags of 3 colors that can be made by selecting from the 6 available colors.

N ° of 3-color flags = 6P3 = 6! / (6 - 3)!

N ° of 3-color flags = 6 * 5 * 4 = 120 flags

b) You want to find the number of different flags of 3 colors that can be made by selecting from the 8 available colors.

N ° of 3-color flags = 8P3 = 8! / (8 - 3)!

N ° of 3-color flags = 8 * 7 * 6 = 336 flags

c) The number of different 4-color flags that can be made by selecting from the 8 available colors must be calculated.

Number of 4-color flags = 8P4 = 8! / (8 - 4)!

No. of 4-color flags = 8 * 7 * 6 * 5 = 1680 flags

d) You want to determine the number of different 2-color flags that can be made by selecting from the 8 available colors.

N ° of 2-color flags = 8P2 = 8! / (8 - 2)!

No. of 2-color flags = 8 * 7 = 56 flags

References

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