# How many permutations can be formed by sampling 5 items from 6 without replacement?

• Last Updated : 29 Nov, 2021

Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

Permutation Formula

In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

nPr = (n!)/(n – r)!

Here,

n = group size, the total number of things in the group

r = subset size, the number of things to be selected from the group

Combination

A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Combination Formula

In combination r things are picked from a set of n things and where the order of picking does not matter.

nCr =n! ⁄ ((n-r)! r!)

Here,

n = Number of items in set

r = Number of things picked from the group

### How many permutations can be formed by sampling 5 items from 6 without replacement?

Solution:

Consider the question above, but repetition is not allowed. if A={1,2,3,4,5,6} and k=5, there are 720  different possibilities:

In common, we can say that there are k place in the selected list:

(place 1, place 2, …, place k). There are n options for the first place , (n−1)

options for the second place (since one component has already been allocated to the

first place and cannot be chosen here), (n−2) options for the third place, … (n−k+1)

options for the kth place. Thus, when ordering matters and repetition is not allowed,

the total number of ways to choose k objects from a set with n component is n×(n−1)×…×(n−k+1).

Any of the selected lists in the above setting (select k component, ordered and no repetition) is called a k-permutation of the component in set A. We use the following notation to show the number of k-permutations of an n-component set:

nPk = n×(n−1)×…×(n−k+1).

Now, we have 720 possibility, the permutation can be formed by sampling 5 items from 6.

like A = {1,2,3,4,5,6}

1.(1,2,3,4,5)  11.(1,2,4,5,6) 21.(1,2,6,3,5)  31.(1,3,4,2,5) 41.(1,3,5,4,6)

2.(1,2,3,5,4)  12.(1,2,3,6,5) 22.(1,2,6,5,3)  32.(1,3,4,5,2) 42.(1,3,5,6,4)

3.(1,2,3,4,6)  13.(1,2,5,3,6) 23.(1,2,6,4,5)  33.(1,3,4,2,6) 43.(1,3,6,2,4)

4.(1,2,3,6,4)  14.(1,2,5,6,3) 24.(1,2,6,5,4)  34.(1,3,4,6,2) 44.(1,3,6,4,2)

5.(1,2,3,5,6)  15.(1,2,5,3,4) 25.(1,3,2,4,5)  35.(1,3,4,5,6) 45.(1,3,6,2,5)

6.(1,2,3,6,5)  16.(1,2,5,4,3) 26.(1,3,2,5,4)  36.(1,3,4,6,5) 46.(1,3,6,5,2)

7.(1,2,4,3,5)  17.(1,2,5,4,6) 27.(1,3,2,4,6)  37.(1,3,5,2,4) 47.(1,3,6,4,5)

8.(1,2,4,5,3)  18.(1,2,5,6,4) 28.(1,3,2,6,4)  38.(1,3,5,4,2) 48.(1,3,6,5,4)and so on….

9.(1,2,4,3,6)  19.(1,2,6,3,4) 29.(1,3,2,5,6)  39.(1,3,5,2,6)

10.(1,2,4,6,3) 20.(1,2,6,4,3) 30.(1,3,2,6,5)  40.(1,3,5,6,2)

Likewise when we put 1 on the first position we have 120 possibilities on the same way when we put 2 on the first position we have 120 possibilities same for 3,4,5 and 6.

So, when we do the total of all the possibilities we get = 120+120+120+120+120+120

= 720

So, 720 permutations can be formed by sampling 5 items from 6 without replacement

### Similar Questions

Question 1: What is the birthday problem?

Let us consider this example to understand birthday problem

There are total 30 people in the room

What is the probability that at least two people have the same birthday or what is the probability that someone in the room share His/Her birthday with at least someone else

White color = p(someone shares with at least                                someone else) or p(s)                         Green color = p(all 30 people have different            birthday) or p(d)

p(s) + p(d) = 1 or 100%

p(s) = 100% – p(d)

If we have two person:

Lets see, person one, their birthday could be 365 days out of 365 days

Now person two could be born on any day that person was not born on

So, 365⁄365 (1st person birthday) 364⁄365 (2nd person birthday)

= 365 × 364 ⁄ 365² = (365! ⁄ (365-2)!) ⁄ 365² = (365! ⁄ 363!) ⁄ 365²

If we have three person:

So, 365⁄365 (1st person birthday) 364⁄365 (2nd person birthday) 363⁄365 (3rd person)

= 365 × 364 × 363 ⁄ 3653 = (365! ⁄ (365-3)!) ⁄ 3653 = (365! ⁄ 362!) ⁄ 3653

Now, if we have 30 people, the probability that no one shares the same birthday

= 365× 364×363×……×336 ⁄ 36530 = (365! ⁄ (365-30)!) ⁄ 36530 = (365! ⁄ 335!) ⁄ 36530 = .2936 or 29.36%

p(d) = .2936 or 29.36%

p(s) = 100% – p(d)

= 100% – 29.36% or 1 – .2936

= .7063 ≈ 70.6%

Question 2: How many permutations can be formed by sampling 3 items from 4 without replacement?

Solution:

Consider the question above, but repetition is not allowed. if A={1,2,3,4} and k=3, there are 24 different possibilities:

In common, we can say that there are k place in the selected list:

(place 1, place 2, …, place k). There are n options for the first place, (n−1) options for the second place (since one component has already been allocated to the first place and cannot be chosen here), (n−2) options for the third place, … (n−k+1) options for the kth place. Thus, when ordering matters and repetition is not allowed,

The total number of ways to choose k objects from a set with n component is

n×(n−1)×…×(n−k+1).

Any of the selected lists in the above setting (select k component, ordered and no repetition) is called a k-permutation of the component in set A. We use the following notation to show the number of k-permutations of an n-component set:

nPk = n×(n−1)×…×(n−k+1)

Now, we have 24 possibility, the permutation can be formed by sampling 3 items from 4.

like A = {1,2,3,4}

1.(1,2,3)              11.(2,4,3)               21.(4,1,3)

2.(1,3,2)              12.(2,3,4)               22.(4,3,1)

3.(1,2,4)              13.(3,1,2)               23.(4,2,3)

4.(1,4,2)              14.(3,2,1)               24.(4,3,2)

5.(1,3,4)              15.(3,1,4)

6.(1,4,3)              16.(3,4,1)

7.(2,1,3)              17.(3,2,4)

8.(2,3,1)              18.(3,4,2)

9.(2,1,4)              19.(4,1,2)

10.(2,4,1)             20.(4,2,1)

Likewise when we put 1 on the first position we have 6 possibilities, in the same way when we put 2 on the first position we have 6 possibilities same for 3 and 4.

So, when we do the total of all the possibilities we get = 6+6+6+6

= 24

SO, 24 permutations can be formed by sampling 3 items from 4 without replacement

Question 3: How many permutations can be formed by sampling 4 items from 5 without replacement?

Solution:

Consider the question above, but repetition is not allowed. if A={1,2,3,4,5} and k=4, there are 120  different possibilities:

In common, we can say that there are k place in the selected list:

(place 1, place 2, …, place k). There are n options for the first place , (n−1) options for the second place (since one component has already been allocated to the first place and cannot be chosen here), (n−2) options for the third place, … (n−k+1) options for the kth place.

Thus, when ordering matters and repetition is not allowed, the total number of ways to choose k objects from a set with n component is

n×(n−1)×…×(n−k+1).

Any of the selected lists in the above setting (select k component, ordered and no repetition) is called a k-permutation of the component in set A. We use the following notation to show the number of k-permutations of an n-component set:

nPk = n×(n−1)×…×(n−k+1).

Now, we have 120 possibilities, the permutation can be formed by sampling 4 items from 5.

like A = {1,2,3,4,5}

1.(1,2,3,4)              11.(1,3,4,5)               21.(1,5,2,4)

2.(1,2,4,3)              12.(1,3,5,4)               22.(1,5,4,2)

3.(1,2,3,5)              13.(1,4,2,3)               23.(1,5,3,4)

4.(1,2,5,3)              14.(1,4,3,2)               24.(1,5,4,3) and so on….

5.(1,2,4,5)              15.(1,4,2,5)

6.(1,2,5,4)              16.(1,4,5,2)

7.(1,3,2,4)              17.(1,4,3,5)

8.(1,3,4,2)              18.(1,4,5,3)

9.(1,3,2,5)              19.(1,5,2,3)

10.(1,3,5,2)             20.(1,5,3,2)

Likewise when we put 1 on the first position we have 24 possibilities on the same way when we put 2 on the first position we have 24 possibilities same for 3, 4 and 5.

so, when we do the total of all the possibilities we get = 24+24+24+24+24

= 120

So, 120 permutations can be formed by sampling 4 items from 5 without replacement.

My Personal Notes arrow_drop_up