How many 4 - letter words (with or without meaning) containing two vowels can be constructed using only the letters (without repetition) of the word 'LUCKNOW'?This question was previously asked in Show
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Answer (Detailed Solution Below)Option 1 : 240 Free Electric charges and coulomb's law (Basic) 10 Questions 10 Marks 10 Mins Concept: If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time. It can be represented as nPr = \(\frac{n!}{(n-r)!}\). The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” nCr = \(\frac{n!}{r!(n-r)!}\) , when n < r Where n = distinct object to choose from C = Combination r = spaces to fill Calculation: Vowels = 2 Consonants = 5 Total Alphabets = 7 Since 4 letter words must include 2 vowels, we don't need to select them, and the rest of the 2 letters will be taken from 5 consonants. Number of ways of selecting 2 letters from 5 consonants = 5C2 = 10 Arrangement of all 4 letters will be given by 4! = 24 ways Total number of arrangements = 5C2 × 4! = 10 × 24 = 240 ways ∴ The total number of words that can be formed is 240. Last updated on Sep 16, 2022 Union Public Service Commission (UPSC) has released the admit card for the UPSC NDA II 2022 exam. The Prelims exam of NDA II 2022 is scheduled to be held on 4th September 2022. The last date to download the admit card will be 4th September 2022. A total number of 400 vacancies are released for the UPSC NDA II 2022 exam. The selection process for the exam includes a Written Exam and SSB Interview. Candidates who get successful selection under UPSC NDA II will get a salary range between Rs. 15,600 to Rs. 39,100. Know the UPSC NDA preparation strategy here. In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N. Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be `""^2P_2 = 2!` Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time. Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440 Table of Contents Nội dung chính
Therefore, 1440 words with or without meaning, can be formed using all the letters of the word ‘EQUATION’, at a time so that the vowels and consonants occur together. How many words can be formed with the letters of the word pataliputra without changing the relative positions of vowels and consonants?Therefore, the number of words that can be formed with the letters of the word $\text{PATALIPUTRA}$ without changing the relative positions of vowels and consonants is 3600. How many words with or without meaning can be formed by using the letters of the word mixture so that the vowels are never together? How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together? Therefore, total no. of words =2×120×6=1440. How many words can be formed with the letters of the word triangle so that vowels occupy odd places?Hence, the number of words where vowels occupy odd places are 576. How many different words can be formed with the letters of the EQUATION so that 1 the words begin with E and end with N 2 the words begin and end with a consonant?Step-by-step explanation: 6×720=4320. How many words can be formed by using all letters of Tihar? Answer is “120” How many words can be formed by using letters of the word Delhi?5 different How many words can be formed from the letters of the word director?Hence the total number of words which can be formed using the letters of the word DIRECTOR =8! How many words with or without meaning can be formed by using all the letters of word refund if repetition of letters is not allowed? If repetition is allowed, the number of words we can form = 4*4*4*4*4 = 1024. (This is because, when repetition is allowed, we can put any of the four unique alphabets at each of the five positions.) If repetition is not allowed, the number of words we can form = 5!/2! = 60. How many 3 letter words with or without meaning can be formed out of the letters of the word hexagon if repetition of letters is not allowed?= 120 ways. The 3 vowels can be arranged among themselves in 3! How many words can be made from the letters of the word Delhi?Sample Problems. Question 1 : How many words can be formed by using 3 letters from the word “DELHI”? Solution : The word “DELHI” has 5 different words. How many words can be formed from the letter of the word TRIANGLE how many of them have the word angle? No. of words that can be made using the letter “TRIANGLE” is 8!= 40320 ways. How many letters are there in the equation Equation?The word ‘EQUATION’ has 8 letters. It has 5 vowels (A E I O U) AND 3 consonants (Q T N) in it and these 5 vowels should always come together AND 3 consonants come together. letter. That is, (A E I O U). letter. That is, (Q T N). Hence we can assume total letters as 2 and all these letters are different. How many vowels are there in the word equation?In the word ‘equation’, there are five vowels (a,e, i, o, u) and three consonants (q, t, n). Since the vowels and consonants should always occur together, we have to group them and consider as two separate things, say V and C. When do vowels and consonants always come together? The question is vowels and consonants should always be together. Hence, consider all vowels as one letter and all consonants as one letter; then we get word of two letters which can be arranged in: So, the final answer when all vowels and consonants should come together is: = 2 *… How many ways can two words be arranged?EQUATION has 8 letters. Since you want constants and vowels together so there will be two groups in the complete words (constants and vowels). So these two words can be arranged in 2! ways= 2*1=2 Now about Vowles- We can arrange 5vowles among themselves in 5! ways. https://www.youtube.com/watch?v=IeVI-1Fg5HU How many different words can be formed with the letters of the word EQUATION so that the word begins with E and ends with N?Hence, we can form a total 4320 different words ending and beginning with consonants with the letters of the word EQUATION. How many different words can be formed with the word EQUATION?Summary: The number of words, with or without meaning, that can be formed using all the letters of the word EQUATION, using each letter exactly once is 40,320. How many different words can be formed with the letters of the word EQUATION the word begin with E?Step-by-step explanation: 6×720=4320. How many different arrangements can be made by using all the letters of Theword EQUATION beginning with vowel and ending with a consonant?The word "Equation" has 8 letters all of which are unique. For the letters in between the end letters, there are 6 of them (5 vowels and the one consonant we didn't use) and can be placed anywhere. That's 6! =720 . How many words with or without meaning can be formed using all the letters of the word EQUATION at a time?Therefore, 1440 words with or without meaning, can be formed using all the letters of the word 'EQUATION', at a time so that the vowels and consonants occur together. How many words with or without meaning can be formed using all the letters of the word Sputnik such that the terminal letters can be vowels only?= 40320. Was this answer helpful? How many different words with or without meaning can be made using all the vowels at a time so that the word does not begin with a?= 120` <br> No. How many different words with or without meaning can be made using all the vowels at a time ?`?1 Answer. There are 5 vowels in 26 alphabets. Hence, using all 5 vowels at a time, number of different words (with or without meaning) can be made are = 5! How many words can be formed without changing the relative positions of the vowels and consonants of the word calculator?Therefore, the number of words that can be formed with the letters of the word $\text{PATALIPUTRA}$ without changing the relative positions of vowels and consonants is 3600. Hence, option (C) is correct. permutation.
How many ways can the letters of the word algebra be arranged without changing the relative order of the vowels and consonants?Therefore in 24 ways the consonants are arranged in the word ALGEBRA. Therefore in 72 ways the letters of the word 'ALGEBRA' are arranged without changing the order of vowels and consonants. So, the correct answer is “72 ways”.
How many words can an EQUATION form?Summary: The number of words, with or without meaning, that can be formed using all the letters of the word EQUATION, using each letter exactly once is 40,320.
How many words with or without meaning EQUATION so that vowels and consonants occur together?Therefore, 1440 words with or without meaning, can be formed using all the letters of the word 'EQUATION', at a time so that the vowels and consonants occur together.
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