How many arrangements can be made from the word probability when all are taken at a time

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, the 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

In how many ways can the letters of the word IMPOSSIBLE be arranged so that all the vowels come together?

Solution:

Vowels are: I,I,O,E

If all the vowels must come together then treat all the vowels as one super letter, next note the letter ‘S’ repeats so we’d use

7!/2! = 2520 

Now count the ways the vowels in the super letter can be arranged, since there are 4 and 1 2-letter(I’i) repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

= (7!/2! × 4!/2!) 

= 2520(12)

= 30240 ways

Similar Questions

Question 1: In how many ways can the letters be arranged so that all the vowels came together word is CORPORATION?

Solution:

Vowels are :- O,O,A,I,O

If all the vowels must come together then treat all the vowels as one super letter, next note the R’r letter repeat so we’d use

7!/2! = 2520

Now count the ways the vowels in the super letter can be arranged, since there are 5 and 1 3-letter repeat the super letter of vowels would be arranged in 20 ways i.e., (5!/3!)

= (7!/2! × 5!/3!)

= 2520(20)

= 50400 ways

Question 2: In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together?

Solution:

Vowels are :- A,A,E,I

Next, treat the block of vowels like a single letter, let’s just say V for vowel. So then we have MTHMTCSV – 8 letters, but 2 M’s and 2 T’s. So there are

8!/2!2! = 10,080

Now count the ways the vowels letter can be arranged, since there are 4 and 1 2-letter repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

= (8!/2!2! × 4!/2!)

= 10,080(12)

= 120,960 ways

Question 3: In How many ways the letters of the word RAINBOW be arranged in which vowels are never together?

Solution:

Vowels are :- A, I, O  

Consonants are:- R, N, B, W.

Arrange all the vowels in between the consonants so that they can not be together. There are 5 total places between the consonants. So, vowels can be organize in 5P3 ways and the four consonants can be organize in 4! ways.

Therefore, the total arrangements are 5P3 * 4! = 60 * 24 = 1440

Result

Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. A typical combination lock for example, should technically be called a permutation lock by mathematical standards, since the order of the numbers entered is important; 1-2-9 is not the same as 2-9-1, whereas for a combination, any order of those three numbers would suffice. There are different types of permutations and combinations, but the calculator above only considers the case without replacement, also referred to as without repetition. This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example, 3-3-3.

Permutations

The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as 3-3-3. For example, in trying to determine the number of ways that a team captain and goalkeeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goalkeeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:

A B C D E F G H I J K   11 members; A is chosen as captain

B C D E F G H I J K   10 members; B is chosen as keeper

As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goalkeeper, A was removed from the set before the second choice of the goalkeeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 × 10 × 9 × 8 × 7 × ... × 2 × 1, or 11 factorial, written as 11!. However, since only the team captain and goalkeeper being chosen was important in this case, only the first two choices, 11 × 10 = 110 are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 × ... × 2 × 1, or 9!. Thus, the generalized equation for a permutation can be written as:

Or in this case specifically:

Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:

nPr = nr

Combinations

Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways, including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply

. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of 11. Unlike the case given in the permutation example, where the captain was chosen first, then the goalkeeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters A through K, it does not matter whether A and then B or B and then A are chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all n people, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations (110 from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, A then B or B then A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:

Or in this case specifically:

It makes sense that there are fewer choices for a combination than a permutation, since the redundancies are being removed. Again for the curious, the equation for combinations with replacement is provided below:

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