How many words , with or without meaning can be formed with letters of the word "EQUATION" at a time so that Vowels and Consonents occur together ?

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Let us say C = group of consonants and V = group of vowels. For these groups only two cases of arrangement arise CV or VC There are 3 consonants and 5 vowels in the word EQUATION. Under group C there are 3! choices of arrangement within themselves. and under group V, there are 5! choices...
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Let us say C = group of consonants and V = group of vowels. For these groups only two cases of arrangement arise CV or VC There are 3 consonants and 5 vowels in the word EQUATION. Under group C there are 3! choices of arrangement within themselves. and under group V, there are 5! choices of arrangement within themselves. So, total choices of arrangement of letters according to the stated criterion that all vowels are grouped together and all consonants together. = 2 *3!*5! = 2 * 6 * 120 = 1440 read less
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there are 5 vowels and 3 consonants. 2 groups formed, one with vowels and other with consonants. So those 2 groups can be arranged in 2! ways. Now vowels in themselves can be arranged in 5! ways and consonants in themselves can be arranged in 3! ways. So totally we have 2! * 5! * 3! ways i.e., 2*1*5*4*3*2*1*3*2*1...
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there are 5 vowels and 3 consonants. 2 groups formed, one with vowels and other with consonants. So those 2 groups can be arranged in 2! ways. Now vowels in themselves can be arranged in 5! ways and consonants in themselves can be arranged in 3! ways. So totally we have 2! * 5! * 3! ways i.e., 2*1*5*4*3*2*1*3*2*1 = 1440 ways read less
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The word EQUATION consists of 5 vowels and 3 consonants. 5 words can be arranged in 5! = 120 ways , 3 consonants can be arranged in 3! = 6 ways . vowels and consonants can be arranged in 2! = 2 ways . therefore, total number of words can be formed = 120 X 6 X 2 = 1440
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5!x3!x2!
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Vowels= A, E, I, O, U =5 Consonants= Q, T, N = 3 Arranging Vowels = 5! ways Arranging Consonants = 3! ways But Vowels or Consonants may come first, which is of 2 ! ways Total arrangements= 2! x 3! x 5! ways.
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The word EQUATION consists of 8 different letters.Out of which E,U,A,I,O are vowels and Q,T,N are consonants. In the required words, we tie the vowels together and the consonants together. These two groups can be arranged in 2! ways .Corresponding to each of these arrangements , the 5 vowels can...
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The word EQUATION consists of 8 different letters.Out of which E,U,A,I,O are vowels and Q,T,N are consonants. In the required words, we tie the vowels together and the consonants together. These two groups can be arranged in 2! ways .Corresponding to each of these arrangements , the 5 vowels can be arranged in 5! ways and 3 Consonants can be arranged in 3! ways. hence total number of words = 2*(5!)*(3!)= 1440 read less
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Let us consider all the vowels as single letter say X, and all the consonants as other single letter Y , now these two letters X and Y can shuffle in 2! Ways. but 5 vowels can interchange their positions in 5! Ways and 3 consonants can interchan ge their positions in 3! Ways, ? total number of...
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Let us consider all the vowels as single letter say X, and all the consonants as other single letter Y , now these two letters X and Y can shuffle in 2! Ways. but 5 vowels can interchange their positions in 5! Ways and 3 consonants can interchan ge their positions in 3! Ways, ? total number of words formed =2!x3!x5! =2x6x120=1440 read less
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factorial 2 * factorial 5 * factorial 3
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EQUATION - contains 5 vowels = 5! ways & 3 Consonants = 3! ways Vowels and consonants can be arranged in 2ways = 2! ways Therefore, 5! X 3! X 2! = 120 X 6 X 2 = 1440
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