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
The number of arrangements that can be formed out of “LOGARITHM” so that(1) No two vowels come together is (A) $6!\,7{p_3}$ (B) $6!\,7!$(C) $6!\,3!$(D) $7!\,3!$(2) All the vowels not come together is (A) $6!\,7{p_3}$(B) $9! - 7!\,3!$(C) $7!\,3!$(D) $6!\,7!$(3) No two consonants come together is (A) $0$(B) $6!\,7!$(C) $6!\,3!$(D) $7!\,3!$
Answer
Verified
Hint: Here, the given question is from permutation and combination. We have to find the number of arrangements that can be formed out by the letters of the word “LOGARITHM” considering different conditions. So, firstly we have to count the number of consonants and vowels in the word “LOGARITHM”, then apply the conditions such as in the first question arrange the consonant by putting a space between two consonants and then arrange the vowels in that place.
Complete step-by-step
solution: (1) Here, the letters of the word “LOGARITHM” are arranged in such a way that no two vowels come together. Since, the given condition is that no two vowels come together and total six consonants and three vowels are in the word “LOGARITHM” so the possible arrangement may be VCVCVCVCVCVCV. And of the six positions of consonant, six consonants can be arranged in $6!$ ways. The arrangement of six consonants produces seven places for vowels and we have only three vowels so this
can be arranged as $7{p_3}$. So, the total possible number of arrangements is $6!\,7{p_3}$. Hence, option (A) is correct. (2) Here, the letters of the word “LOGARITHM” are arranged in such a way that all the vowels do not come together. To arrange according to given condition, find the total possible number of arrangements of letters without any condition that is $9!$ and then subtract the case when all the vowels come together from this. Now, put all vowels in to a box and
consider it as a single letter and make arrangements of six consonants and a box of vowels that comes as $7!\,3!$ because the seven letters can be arranged in $7!$ and the vowels inside the box can be arranged internally in $3!$ ways. So, the total possible number of arrangements is $9! - 7!\,3!$ Hence, option (B) is correct. (3) Here, the letters of the word “LOGARITHM” are arranged in such a way that no two consonants come together. We have only three vowels but six
consonants and we have to arrange such that no two consonants come together and this can be possible if we arrange the letters as CVCVCVC but there is only four places for consonants but we have to arrange six consonants so, it is not possible to arrange in such a way. So, the possible number of arrangements is zero. Hence, option (A) is correct.
Note: We should know, $n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\,}}$ \[n{C_r} = \dfrac{{n!}}{{\left( {n - r}
\right)!\,r!}}\] It will be helpful in finding the numerical value.
How many ways can the letter of the word section be arranged such that no two vowels are together?
Answer: The number of ways in which letters of SECTION are arranged such that that no vowels are together are 1440 ways.
How many words can be formed such that no two vowels come together?
The total amount of arrangements with no two vowels together is trickier. So, there are 1152 combinations, or 22.9% of all the combinations.
How many different ways can the letters of the word combine be arranged so that the vowels always come together?
The number of ways the word TRAINER can be arranged so that the vowels always come together are 360.
How many arrangements are there where no two vowels are next to each other?
