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How many words can be formed, each of $2$ vowels and $3$ consonants from letters of the word "DAUGHTER" What my textbook has done: it has first taken combinations of vowels and then consonants then multiplied them altogether. Now for each combination of words they can be shuffled in $5!$ ways, so multipliying by $5!$ we get the required answer. My question is: why has the book used combinations instead of permutations while selecting vowels and consonants? Thanks.
 JKnecht 6,2955 gold badges23 silver badges44 bronze badges asked Jun 4, 2015 at 19:00
$\endgroup$ $\begingroup$ All the letters are different, so that makes things easier. Pick the two vowels ($_3C_2$) and pick the three consonants ($_5C_3$) and then pick what order they go in $(5!)$. So the answer is $3 \cdot 10 \cdot 120 = 3600.$ You take combinations of the vowels and consonants because the order of them doesn't matter at that point. You order them in the last step, after you've chosen which ones go in your five-letter word. In other words, it doesn't matter that I pick $A$, then $U$, instead of $U$, then $A$. It just matters that I picked the set $(A,U)$. answered Jun 4, 2015 at 19:12
JohnJohn 25.7k3 gold badges35 silver badges59 bronze badges $\endgroup$ 2 $\begingroup$ Your query why not permutation first ? As, you have to make words of length=$5$. And of these $5$, $2$ are vowels and $3$ consonants. Since, you have to first get those $2$ vowels and $3$ consonants to make the desired word. So first operation has to be combination(selection operation), which will select $2$ vowels out of $3$ vowels(A,E,U) and then you have to select 3 consonants out of $5$(D,G,H,T,R). And they need to be multiplied, as there can be many such combinations i.e $C(3,2)*C(5,3)$. Now that you have formed $5$ letter word. These letters can be arranged among themselves to make different words. Hence, you need to apply permutation(arrangement) i.e. $5!$, making final result= $C(3,2)*C(5,3)*5!$. answered Jun 4, 2015 at 19:40
user2016963user2016963 1111 silver badge10 bronze badges $\endgroup$ 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?Nội dung chính
Answer Verified Hint: Permutations are the different ways in which a collection of items can be
arranged. For example: Complete step-by-step answer: 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. Note: Always keep an eye on the keywords used in the question. The
keywords can help you get the answer easily. How many words, with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once?Answer Verified Hint: In order to solve such questions where we need total number of words or indirectly total number of arrangements, use the concept of permutation. Use the formula for counting the number of words. Complete step-by-step answer: \[ = {}^8{P_8} \] \[ = \dfrac{{8!}}{{\left( {8 - 8} \right)!}}\] \[ = \dfrac{{8!}}{{0!}} \] \[ = \dfrac{{8!}}{1}{\text{ }}\left[ {\because 0! = 1} \right] \] Hence, 40320 words with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once. Note: A permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen objects does matter. In other words, a permutation is an arrangement of objects in a definite order. Students must remember different formulas for finding the permutation. And also must take extra care when in such problems the letters are repeating. Solution There are 8 letters in the word EQUATION including 5 vowels and 3 consonants.Now 5 vowels can be arranged in 5! ways and 3 consonants can be arranged in 3! ways. Also the two groups of vowels and consonants can be arranged in 2! ways.∴ Total number of permutations= 5!×3!×2!=120×6×2=1440.How many different words with or without meaning can be formed using all the letters of the word iterate?Hence, the number of words are 120. Q. How many words with or without meaning can be formed using all the letters of the word mathematics?Total no of cases in which the word MATHEMATICS can be written = 11! = 8! Hence, the number of words can be made by using all letters of the word MATHEMATICS in which all vowels are never together is 378000. How many words with or without meaning can be formed using all the letters of the word Monday using each letter exactly once?Therefore, the number of words that can be formed using all the letters of the word MONDAY, using each letter exactly once is 6×5×4×3×2×1=6! =720. How many words with or without meaning can be formed using all the letters of the word Covid?Detailed Solution The correct answer is 120. No of letters in COVID= 5. Now as one letter had already taken place, the number of letters left to be filled at tens place is 4. How many words of 3 consonants and 2 vowels can be formed out of 5 consonants and 3 vowels?Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. = 5! = 120.
How many different words each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17?=6800×120=816000.
How many 5 letter words contain 3 vowels and 2 consonants?1440`. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word mathematics?Therefore, 30 words can be formed from the letters of the word DAUGHTER each containing 2 vowels and 3 consonants. Note: A Permutation is arranging the objects in order.
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