Answer
Hint: First find the number of ways in which word ‘Mathematics’ can be written, and then we use permutation formula with repetition which is given as under,
Number of permutation of $n$objects with$n$, identical objects of type$1,{n_2}$identical objects of type \[2{\text{ }} \ldots \ldots .,{\text{ }}{n_k}\]identical objects of type $k$ is \[\dfrac{{n!}}{{{n_1}!\,{n_2}!.......{n_k}!}}\]
Complete step by step solution:
Word Mathematics has $11$ letters
\[\mathop
{\text{M}}\limits^{\text{1}} \mathop {\text{A}}\limits^{\text{2}} \mathop {\text{T}}\limits^{\text{3}} \mathop {\text{H}}\limits^{\text{4}} \mathop {\text{E}}\limits^{\text{5}} \mathop {\text{M}}\limits^{\text{6}} \mathop {\text{A}}\limits^{\text{7}} \mathop {\text{T}}\limits^{\text{8}} \mathop {\text{I}}\limits^{\text{9}} \mathop {{\text{ C}}}\limits^{{\text{10}}} \mathop {{\text{ S}}}\limits^{{\text{11}}} \]
In which M, A, T are repeated twice.
By using the formula \[\dfrac{{n!}}{{{n_1}!\,{n_2}!.......{n_k}!}}\],
first, we have to find the number of ways in which the word ‘Mathematics’ can be written is
$
P = \dfrac{{11!}}{{2!2!2!}} \\
= \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1 \times 2 \times 1}} \\
= 11 \times 10 \times 9 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\
= 4989600 \\
$
In
\[4989600\]distinct ways, the letter of the word ‘Mathematics’ can be written.
[i] When vowels are taken together:
In the word ‘Mathematics’, we treat the vowels A, E, A, I as one letter. Thus, we have MTHMTCS [AEAI].
Now, we have to arrange letters, out of which M occurs twice, T occurs twice, and the rest are different.
$\therefore $Number of ways of arranging the word ‘Mathematics’ when consonants are occurring together
$
{P_1} = \dfrac{{8!}}{{2!2!}} \\
= \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \\
= 10080 \\
$
Now, vowels A, E, I, A, has $4$ letters in which A occurs $2$ times and rest are different.
$\therefore $Number of arranging the letter
\[
{P_2} = \dfrac{{4!}}{{2!}} \\
= \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1}} \\
= 12 \\
\]
$\therefore
$Per a number of words $ = [10080] \times [12]$
In which vowel come together $ = 120960$ways
[ii] When vowels are not taken together:
When vowels are not taken together then the number of ways of arranging the letters of the word ‘Mathematics’ are
$
= 4989600 - 120960 \\
= 4868640 \\
$
Note: In this type of question, we use the permutation formula for a word in which the letters are repeated. Otherwise, simply solve the question by counting the number of letters of the word it has and in case of the counting of vowels, we will consider the vowels as a single unit.
Answer
Verified
Hint:Here, we will proceed by observing all the letters in the word MATHEMATICS that are repeating and then, we will use the formula i.e., Permutation of n items out of which x items, y items and z items of different types are repeating = $\dfrac{{n!}}{{x!y!z!}}$. For the next two parts, we will fix the first letter of the word as C and T in order to find out the different arrangements possible.Complete step-by-step answer:
The word MATHEMATICS consists of 2 M’s, 2 A’s, 2 T’s, 1
H, 1 E, 1 I, 1 C and 1 S.
Total number of letters in the word MATHEMATICS = 11
As we know that
Total number of different arrangements of n items out of which x items, y items and z items of different types are repeating = $\dfrac{{n!}}{{x!y!z!}}{\text{ }} \to {\text{[1]}}$
Using the formula given by equation [1], we can write
Total number of different arrangements which can be made by using all the 11 letters in the word MATHEMATICS in which letter M, letter A and letter T are
repeating twice = $\dfrac{{11!}}{{2!2!2!}} = \dfrac{{11.10.9.8.7.6.5.4.3.2!}}{{2.1.2.1.2!}} = \dfrac{{11.10.9.8.7.6.5.4.3}}{4} = 4989600$
Therefore, a total of 4989600 words can be formed using all the letters of the word MATHEMATICS.
For the words which begin with letter C formed using all the letters of the word MATHEMATICS, the first letter is fixed as C so the next 10 letters need to be selected from the left letters [i.e., 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I and 1 S]
Using the
formula given by equation [1], we can write
Total number of different arrangements which can be made by using all the left 10 letters [except letter C] in the word MATHEMATICS in which letter M, letter A and letter T are repeating twice = $\dfrac{{10!}}{{2!2!2!}} = \dfrac{{10.9.8.7.6.5.4.3.2!}}{{2.1.2.1.2!}} = \dfrac{{10.9.8.7.6.5.4.3}}{4} = 453600$
Therefore, a total of 453600 words which begin with C can be formed using all the letters of the word MATHEMATICS.
For the words which
begin with letter T formed using all the letters of the word MATHEMATICS, the first letter is fixed as T so the next 10 letters need to be selected from the left letters [i.e., 2 M’s, 2 A’s, 1 T, 1 H, 1 E, 1 I, 1 C and 1 S]
Also we know that
Total number of different arrangements of n items out of which x items and y items of different types are repeating = $\dfrac{{n!}}{{x!y!}}{\text{ }} \to {\text{[2]}}$
Using the formula given by equation [2], we can write
Total number of
different arrangements which can be made by using all the left 10 letters [except one of the two letters T] in the word MATHEMATICS in which letter M, letter A are repeating twice = $\dfrac{{10!}}{{2!2!}} = \dfrac{{10.9.8.7.6.5.4.3.2!}}{{2.1.2!}} = \dfrac{{10.9.8.7.6.5.4.3}}{2} = 907200$
Therefore, a total of 907200 words which begin with T can be formed using all the letters of the word MATHEMATICS.
Note- In this particular problem, since we have to rearrange the letters of the word MATHEMATICS that’ s why we are using permutation formulas. If we were asked for selection of some letters out of all the letters we would have used combinations formula. The general formula for arrangement of r items out of n items is given by \[{}^n{P_r} = \dfrac{{n!}}{{\left[ {n - r} \right]!}}\].