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
Solution : [a]Number of 8 different arrangements with all vowels occur together-
Vowels include-[a,u,e] so mark them as 1
now we have 6 things to arrange , that can be done in `6! xx 3! =4320` Mulitplying by 3! for internal arrangements of vowels.
[b] all vowels do not occur together
=>Total arrangements-[when vowels occur together]
=>`8!-4320=36000`
Solution : The given word, 'DOUGHTER' contains 3 vowels A, U, E and 5 consonants D, G, H, T, R.
Case [i] When all vowels occur together:
Let us assume [AUE] as a single letter.
Then, this letter [AUE] along with 5 other letters can be arranged in `""^[6]P_[6]=[6!] "ways"=[6xx5xx4xx3xx2xx1]` ways
`=720` ways.
These 3 vowels may be arranged among themselves in `3! = 6` ways.
Hence, the required
number of words with vowels together
`[6!]xx[3!]=[720xx6]=4320.`
Case [ii] When all vowels do not occur together:
Number of words formed by using all the 8 letters of the given word
`=""^[8]P_[8]=8! =[8xx7xx6xx5xx4xx3xx2xx1]=40320.`
Number of words in which all vowels are never together
= [total number of words ] - [ number of words with all vowels together]
`=[40320 - 4320]=36000.`