How many different arrangements can be formed from the letters of the word EQUATION if each arrangement begins and ends with a consonant?
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√ many different arrangements can be formed from the letters of the word, EQUATION if each arrangement begins and ends with a consonant? A.720 B. 2,040 C.3,100 D. 4,320 atmr hocksQuestion  Gauthmathier9846Grade 9 · 2021-04-20 YES! We solved the question! Check the full answer on App Gauthmath √ many different arrangements can be formed from the letters of the word, EQUATION if each arrangement begins and ends with a consonant? Gauthmathier9792Grade 9 · 2021-04-20 Answer Explanation Thanks (60) Does the answer help you? Rate for it!
B.24 ?. How many different arrangements can be formed from the letters of the word, EQUATION if each arrangement begins and end s with a consonant? A.720 C. 3,100 B. 2,040 D. 4,320Question Gauthmathier9541Grade 9 · 2021-04-28 YES! We solved the question! Check the full answer on App Gauthmath B.
B.24 ?. How many different arrangements can be for - Gauthmath ?. How many different arrangements can be formed from the letters of the word, EQUATION if each arrangement begins and end s with a consonant?
Gauthmathier1370Grade 9 · 2021-04-28 Answer Explanation Thanks (172) Does the answer help you? Rate for it!
Answer: A. 720 Step-by-step explanation: 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 Pa Brainliests answer po plsss The word "Equation" has 8 letters all of which are unique. We want the number of letter arrangements that start and end with a consonant. Let's first see that with letter arrangements, we're working with permutations (we care about the order of things). The general formula is: #P_(n,k)=(n!)/((n-k)!); n="population", k="picks"# First let's work with the end letters. We only want consonants (there are 3 of them), which gives: #P_(3,2)=(3!)/(1!)=6# 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#. In total then, we have #6xx720=4320# How many different arrangements can be made out of the letters of the word EQUATION?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 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 formed from the word EQUATION if in each arrangement all vowels are together?1 Answer. James L. Mathematic can be arranged in 453,600 different ways if it is ten letters and only use each letter once. Assuming all vowels will be together 15,120 arrangements.
How many five letter words all different can be formed using the letters of the word EQUATION which ends with N?We know that nPr=n! (n−r)! Therefore , 15120 five letter words can be forms with the letters of the word EQUATIONS without repetition.
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