How many different ways can the letters of the word EXAMINATION be arranged so that vowels always occupy odd places?

In how many ways can the letters of the word 'STRANGE' be arranged so thati the vowels come together?ii the vowels never come together? andiii the vowels occupy only the odd places?

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Text Solution

Solution : The given word consists of 8 letters, out of which there arr 4 vowels and 4 consonants. Let us mark out the positions to be filled up, as shown below: `(""^(1))(""^(2))(""^(3))(""^(4))(""^(5))(""^(6))(""^(7))(""^(8)).` Since vowels occupy odd places, they may be placed at 1, 2, 3, 5, 7. Number of ways of arranging 4 vowels at 4 odd places `= ""^(4)P_(4)=4! =24.` The remaining 4 letters of the given word are consonants, which can be arranged at 4 even places marked 2, 4, 6, 8. Number of ways of arranging 4 consonants at 4 even places `= ""^(4)P_(4)=4! =24.` Hence, the total number of words in which the vowels occupy odd places `=(24xx24) =576.`

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Hint: To find the number of arrangements of letters in a word which has repeating letters can be found out by using the formulae= $ \dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!}} $ Where, n = Total letters to arrangeAnd, p1, p2, p3 = Numbers of repeating letters

Complete step-by-step answer:

Case (i): The word will start with the letter M. Since the word will start with the letter M, the arrangement of letters will beM _ _ _ _ _ _ _ _ _ _ Total number of letters = 11 Now, the number of letters we have to arrange are (11 – 1) = 10We need to arrange letters E, X, A, I, N, A, T, I, O, NNow, we have 1E, 1X, 2A, 2I, 2N, 1T, 1OThe repeating letters are:2A, 2I and 2NSince the letters are repeating, we will be using the formulae = $ \dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!}} $ Since, total number of letters to arrange (n) = 10And, p1 = 2, p2 = 2, p3 =2After putting the values of n, p1, p2, p3 Number of arrangements will be equal to: = $ \dfrac{{10!}}{{2!2!2!}} $ = $ \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}}{{2! \times 2! \times 2!}} $ = $ \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}}{{2 \times 2}} $ = $ 18,14,400 $ So, there are $ 18,14,400 $ ways in which the word ‘EXAMINATION’ can be arranged by keeping the first letter as ‘M’.Case (ii): All the vowels occur together There are 6 vowels in the word ‘EXAMINATION’Since all the vowels have to come together we are going to treat them as single letters. We treat them as [EAIAIO] single objects.Our letter will become:[EAIAIO] _ _ _ _ _ Total number of ways of arranging the 6 vowels together is: Total number of vowels = Total numbers of letters to arrangei.e. n = 6Since, we have 2A and 2I as repeating letters. We are going to use the formulae:= $ \dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!}} $ Putting the values in the formulae, we get= $ \dfrac{{6!}}{{2! \times 2!}} $ $ $ = $ \dfrac{{6 \times 5 \times 4 \times 3 \times 2!}}{{2! \times 2!}} $ Cancelling out one ‘2!’ from the numerator and denominator and we get= $ \dfrac{{6 \times 5 \times 4 \times 3}}{{2!}} $ Simplifying the above equation, we get= $ 6 \times 5 \times 2 \times 3 $ = $ 180 $ Since the vowels are treated as single letters together but we can arrange them together by 180 ways. After the vowels come together, the resultant word looks like[EAIAIO] _ _ _ _ _Total numbers of letters we have to arrange are:= 1 + 5=6i.e. n = 6 The repeating letters are: 2Ni.e. p = 2 Now using the formulae: $ \dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!}} $ Putting the values in the given formulae, we get= $ \dfrac{{6!}}{{2!}} $ = $ \dfrac{{6 \times 5 \times 4 \times 3 \times 2!}}{{2!}} $ Simplifying the equation, we get= $ 6 \times 5 \times 4 \times 3 $ = $ 360 $ Therefore, the number of ways the word will be arranged = $ 360 $ Hence, the required arrangement is:= $ 180 \times 360 $ = $ 64800 $ Therefore, the total number of ways in which the word ‘EXAMINATION’ can be written while keeping the vowels together = $ 64800 $

Note: Consider the fact that the single letter whose values are (1!) are not written in the simplification processes sometimes because the value of 1! = 1 only. The fact that the entire vowel combination is considered as one single letter is a very important point to remember.