How many ways can the letters of regulation be arranged so that the vowels come at odd places?

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

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If the letters of the word REGULATIONS are arranged at random,what is the probability that there will be exactly 4 letters between R and E? The answer in my book is given as 11!/(9C4 x 4! x6!x2!) .Shouldn't the answer be upside down because 11!=total number of arrangements?

asked Nov 23, 2013 at 8:36

ayushayush

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There are $\binom{11}{2}$ equally likely ways to choose the two positions that will be reserved for the letters R and E. (Note this does not count where R and E as individual letters go, we are just putting reserved signs on the two positions.)

If the two positions have a gap of $4$ between them, there are $6$ places where the leftmost position can be, and then the other position is determined. It follows that our probability is $\frac{6}{\binom{11}{2}}$.

answered Nov 23, 2013 at 8:41

André NicolasAndré Nicolas

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Ans: 6/55

Totally there are 11 letters.

R,E can occupy 6 positions like (1,6) (2,7) (3,8) (4,9) (5,10) (6,11). In this R and E can interchange among themselves.

In remaining positions the 9 other letters can arrange themselves in 9! ways. So total ways where R and E have 4 letters between them = 6 * 2 * 9!

Required probability = (6 * 2 * 9!)/(11!) = 12/110 = 6/55

answered Jul 1, 2014 at 13:25

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You could take $\dfrac{{\binom{9}{4}}\times4!\times2\times6!} {11!}$ and get 6/55 as your answer.

• $\binom{9}{4}$$\times4!$ because you have to choose the four numbers apart from R and E that should be tucked in between R and E, after which you should account for the possible arrangements of those four numbers.
• $2$ because you have to account for R-E and E-R arrangements
• $6!$ because you take R- insert 4 letters - E as a block, this block plus the 5 other remaining numbers can be arranged in 6! ways
• 11! because that shows the possible arrangements of all the eleven letters.

answered Nov 10, 2017 at 10:12

Meera UnniMeera Unni

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Here R E can occupy (1,6),(2,7),(3,8),(4,9),(5,10),(6,11) and vice versa is also possible .so the number of ways to fill this is 6*2 ways. Then remaining 9 letters can fill in 9! ways. So the probability will be 6*2*9!/11! ( Since there are 11 letters in total. By solving this we get 6/55

And:6/55

answered Jun 5, 2018 at 15:25

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