How many 4 letters words with or without meaning can be formed out of the letters of the word Lucknow?
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Solution The number of objects(n), in this case, is 5, as the word SMOKE has 5 alphabets.and r=3, as 3-letter word has to be chosen.Thus, the permutation will be:Permutation (when repetition is allowed) =5×5×5=125It’s a question that just about every English learner has asked: “Are there any English words that have no vowels?” Nội dung chính
The answer to this depends what you mean by “vowel” and “word.” In this article, we explain what vowel means and how English words without vowels can—and do!—exist. We also talk about the examples that some people may or may not believe are actually words. Are there words with no vowels?There are two things we mean by the word vowel: a speech sound made with the vocal tract open, or a letter of the alphabet standing for a spoken vowel. Cwm and crwth do not contain the letters a, e, i, o, u, or y, the usual vowels (that is, the usual symbols that stand for vowel sounds) in English. But in those words the letter w simply serves instead, standing for the same sound that oo stands for in the words boom and booth. Dr., nth (as in “to the nth degree”), and TV also do not contain any vowel symbols, but they, like cwm and crwth, do contain vowel sounds. Shh, psst, and hmm do not have vowels, either vowel symbols or vowel sounds. There is some controversy whether they are in fact “words,” however. But if a word is “the smallest unit of grammar that can stand alone as a complete utterance, separated by spaces in written language and potentially by pauses in speech,” then those do qualify. Psst, though, is the only one that appears in the Oxford English Dictionary. WATCH: Words That Are Their Own OppositesHow 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 NDA 02/2021: Maths Previous Year paper (Held On 14 Nov 2021) View all NDA Papers >
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. Solution Total number of letters in word MONDAY = 6Number of vowels in word MONDAY = 2(i) Number of letters used = 4∴ Number of permutations = 6P4=6!(6−4)!=6!2!=6×5×4×3×2!2!=360(ii) Number of letters used = 6∴ Number of permutations = 6P6= 6!0!=6×5×4×3×2×1=720(iii) Here the first letter is vowel.∴ Number of permutation of vowel = 2P1= 2!1!=2 Now the remaining five places can be filled with remaining five letters.∴ Number of permutations = 5P5= 5!0!=5×4×3×2×1=120Thus total number of permutations= 2×120=240.How many different words with or without meaning can be made using all the vowels at a time?=120×6×2=1440. 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, 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 by using all the letters of the word Santiago so that all the vowels come together?= 120` How many words can be formed with or without meaning using all the letters of the word Richard '?= 40320. Was this answer helpful? How many 4So, the total arrangement is given by, 10×9×8×7=5040 .
How many 4Solution : There are 10 letters in the word LOGARITHMS. So, the number of 4-letter words is equal to the number of arrangements of 10 letters, taken 4 at a time, i.e., `. ^(10)P_(4)=5040`. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
How many 4∴ The total number of words that can be formed is 240.
How many 4The number of four-letter words that can be formed using the letters of the word "flash", is 120.
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