
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?
$A)10080$
$B)4989600$
$C)120960$
$D)$ None of these
Answer
475.5k+ views
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.
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.
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