Question

In how many words can the letters of word ‘Mathematics’ be arranged so that (i) vowels are together (ii) vowels are not together

Answer 1

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.