How many arrangements can be made with all the letters of the word mathematics if I all vowels do not occur together?

Answer

Verified

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.

Text Solution

Solution : (i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different. Hence, the total number of arrangements of the given letters `=(11!)/((2!)xx(2!)xx(2!)) =4989600.` (ii) The given word contains 4 vowels AEAI as one letter, we have to arrange 8 letters MATHMTCS + AEAI, out of which M occurs twice, T occurs twice and the rest are all different. So, the number of all such arrangements `=(8!)/((2!)xx(2!))=10080.` Now, out of 4 vowels, A occurs twice and the rest are all distinct. So, the number of arrangements of these vowels `=(4!)/(2!)=12.` Hence, the number of arrangement in which 4 vowels are together `=(10080 xx 12)=120960.`