How many 5 letter words can be formed with 2 vowels and three consonants?
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Solution : There are 5 vowels and 3 consonants in the word 'EQUATION'. Three vowels out of 5 and 2 consonants out of 3 can be chosen in `""^(5)C_(3)xx""^(3)C_(2)` ways. So, there are `""^(5)C_(3)xx""^(3)C_(2)` groups each containing 3 consonants and two vowels. Now, each group contains 5 letters which are to be arranged in such a way that 2 consonats occur together. Considering 2 consonants as one letter we have 4 letters which can be arranged in 4! ways. But two consonants can be put together in 2! ways. Therefore, 5 letters in each group can be arranged in `4!xx2!` ways. 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= $C(3,2)*C(5,3)*5!$. It is right/wrong and better/worse and complete/incomplete and compact/elaborate and ... regardless of who wrote it, when, and why.
I'm working on adapting Algebra solutions to (1) learning style, (2) prerequisite/remedial knowledge, (3) past performance [like computer games track levels], (4) mood-of-the-day, (5) etc. to computerize these "typical" problems. That means graphics/color/animation/sound/etc. and 25 teachers (coordinated, PLZ) to one student rather than 25 students to one teacher. There is now money for educational software!! Care to learn Algebra from IBM's Watson? Average students could learn an entire year of high school Algebra in two weeks!! Report 04/14/17  Kenneth S. answered • 04/14/17 Tutor 4.8 (62)Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018 See tutors like this See tutors like this An interesting problem. First, let's choose two vowels, which can be done in 4C2 = 6 ways. Next, let's choose three consonants, which can be done in 6C3 = 20 ways.
To reflect all possible combinations of two vowels and three consonants, we shall multiply: 120 such choices. But these don't end the problem. Now that we have any particular combination of the vowels & consonants, these five things can be arranged in 5! orders. Step 2: Once the spots for the vowels have been determined, lets count the number of ways we can start filling in with letters: vowel spot 1 - has 5 choices vowel spot 2 - has 5 choices consonant spot 1 - has 21 choices consonant spot 2 - has 21 choices consonant spot 3 - has 21 choices so the 5 letters can be filled in the 5 spots in the following number of ways: 5 * 5 * 21 * 21 * 21 = (5^2 * 21^3) Step 3: Multiply the result from step 1 and step 2 to get the total number of 5-letter words containing two vowels and three consonants So the answer is (5 C 2) * (5^2 * 21^3) = 2315250 b) How many 5-letter words with two vowels and three consonants consist of five different letters, appearing in alphabetical order? Key observation: We need five different letters, appearing in alphabetical order. If you take a specific combination of the 5 different letters, for example, "defabc". Although there are 5! different arrangements, we only need 1 where all letters are arranged in alphabetical order, which is "abcdef". So, effectively, we only need to count how many different combinations of 5 different letter words containing 2 vowels and 3 consonants exist! Page 1: heart, board, death, green, three, mouth, dream, laugh, faith, earth, smile, South, stone, blood, cough, point, phone, grace, knife, train, brain, Eight, fruit, field, dance, royal, cloud, state, bread, beard, place, chair, badge, white, clear, young, great, sweet, beach, space, peach, sound, youth, plate, guard, dough, sleep, whale, James, and brave The largest free 5 letters words list online. Does not include all of the plural forms of five letter words.This should be the most accurate and largest monosyllable words list in the English language.
How many 5 letter words can be formed using 2 vowels and 3 consonants?So now we have 2 vowels and 3 consonants, which means that we have 5 letters in total. Final Answer: Total no. of words formed by using 2 vowels and 3 consonants taken from 4 vowels and 5 constants in equal to 7200.
How many words are formed by 2 vowels and 3 consonants?=60×120=7200.
How many different words containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?So, total number of words = 5C2× 17C3×5! =816000.
How many five letter word sequences consisting of 2 vowels and 3 consonants can be formed from the letters of the word introduce?How many five-letter word sequences consisting of 2 vowels and 3 consonants can be formed from the letters of the word INTRODUCE? 4C2 · 5C3 · 5! = 7200.
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