Hint: There are $11$ letters in the word ‘MATHEMATICS’,
We have to find the number of ways of arranging these letters
Also the given number of letters out of which there are $3$ vowels.
Here, we have to apply the concept of permutation to solve the question.
Finally we get the required answer.Formula used: ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$ Complete step-by-step answer:
As per the question the word ‘Mathematics’ consists of total $11$ letters, out of which $3$ distinct vowels $A,E,A,I$ are present.
So we have $MTHMTCS$ as consonants and $A,E,A,I$ as vowels.
Now we have to arrange $8$letters, $M$ occurs two times and $T$ occurs two times and others are different.
Therefore, the number of ways of arranging these letters$ = $$\dfrac{{8!}}{{2! \times 2!}}$
On splitting the factorial we get,
$ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}$
On multiplying the term we get,
$ = 10080$
Now, we have $4$ letters as vowels $A,E,A,I$ in the word ‘Mathematics’. The vowel $A$ occurs two times and others are distinct.
Here \[n = 4\] and \[r = 2\]
Number of ways of arranging these letters ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$
$ \Rightarrow {}^4{C_2} = \dfrac{{4!}}{{(4 - 2)!}}$
On subtracting the bracket term we get,
$ = \dfrac{{4 \times 3 \times 2!}}{{2!}}$
$ = 12$
The total required number of ways $ = 12 \times 10080$
On multiplying the terms we get
$ = 120960$ Thus the correct option is $C$ Note: In general, permutation can be defined as the act of arranging all the members of a group in an order or sequence. We can also say that if the group is already arranged, then rearranging of the members is known as the procedure of permuting.
Permutation takes place in almost all areas of mathematics. It is specifically used where the order of the data matters.
Combination can be defined as the technique of selecting things from a collection in such a manner that the order of selection does not matter. It is generally used where the order of data does not matter.
In how many ways can the letters of the word IMPOSSIBLE be arranged so that all the vowels come together? Improve Article Save Article Like Article - Last Updated : 29 Nov, 2021
ReadDiscussCoursesPracticeVideoImprove Article Save Article 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.
How many ways the word apple can be arranged so that vowels always come together?
Answer: 60 different ways.
How many ways a letter can be arranged so that vowels come together?
The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.
The number of words which can be made out of the letters of the word MOBILE' when consonants always occupy odd places is \( f \) \( F \) \( F \) \( F \) 72.
How many ways can you arrange the letters of the word mobile such that M is the first letter of the arrangement *?
= 720 ways. Now the Logic behind this arrangement. * The first letter of the arrangement can be any 1 of 6 letters.
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