Answer  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. How many arrangements can be formed of the letters EXAMINATION so that vowels occupy odd places. |
How many arrangements can be formed out of the letter of the word EXAMINATION so that was always occupy odd places?
zusammenhängende Posts
Urheberrechte © © 2024 wiewird Inc.
