How many words with or without meaning can be formed using 2 vowels and 3 consonants from the letters of the word teacher? Mới Nhất

Bí kíp về How many words with or without meaning can be formed using 2 vowels and 3 consonants from the letters of the word teacher? Chi Tiết

Quý quý khách đang tìm kiếm từ khóa How many words with or without meaning can be formed using 2 vowels and 3 consonants from the letters of the word teacher? 2022-10-03 20:38:59 san sẻ Thủ Thuật Hướng dẫn trong nội dung bài viết một cách 2021.

$begingroup$ Nội dung chính How many different words can be formed each containing 2 vowels and 3 consonants? How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word daughter *? How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word honesty *? How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word shoulder? 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 badges45 bronze badges asked Jun 4, năm ngoái 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, năm ngoái at 19:12 JohnJohn 25.7k3 gold badges35 silver badges60 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, năm ngoái at 19:40 user2016963user2016963 1111 silver badge10 bronze badges $endgroup$ How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R. Number of ways of selecting 2 vowels out of 3 vowels =`””^3C_2 = 3` Number of ways of selecting 3 consonants out of 5 consonants = `””^5C_2 = 3` Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30 Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways. Hence, required number of different words = 30 × 5! = 3600 In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R. Number of ways of selecting 2 vowels out of 3 vowels =`””^3C_2 = 3` Number of ways of selecting 3 consonants out of 5 consonants = `””^5C_2 = 3` Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30 Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways. Hence, required number of different words = 30 × 5! = 3600 Concept: Combination   Is there an error in this question or solution? view Video Tutorials For All Subjects Combination video tutorial 00:06:31 Knockout CUET (Physics, Chemistry and Mathematics) Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan. ₹ 7999/- ₹ 4999/- Buy Now Knockout CUET (Physics, Chemistry and Biology) Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan. ₹ 7999/- ₹ 4999/- Buy Now Knockout JEE Main (Six Month Subscription) – AI Coach Study Modules, – Unlimited Mock Tests, – Study Improvement Plan. ₹ 9999/- ₹ 8499/- Buy Now Knockout JEE Main (Nine Month Subscription) – AI Coach Study Modules, – Unlimited Mock Tests, – Study Improvement Plan. ₹ 13999/- ₹ 12499/- Buy Now Knockout NEET (Six Month Subscription) – AI Coach Study Modules, – Unlimited Mock Tests, – Study Improvement Plan. ₹ 9999/- ₹ 8499/- Buy Now How many different words can be formed each containing 2 vowels and 3 consonants? =6800×120=816000. How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word daughter *? Therefore, 30 words can be formed from the letters of the word DAUGHTER each containing 2 vowels and 3 consonants. How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word honesty *? = 5! ×30=120×30=3600. How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word shoulder? Solution 1 Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.

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