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How many different words can be formed with the letters of the word EQUATION without changing the relative order of the vowels and consonants?
  • a)
    125
  • b)
    620
  • c)
    720
  • d)
    880
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
How many different words can be formed with the letters of the word EQ...
To find the number of different words that can be formed with the letters of the word EQUATION without changing the relative order of the vowels and consonants, we need to consider the arrangement of the vowels and consonants separately.

1. Arrangement of Vowels:
The word EQUATION has 4 vowels - E, U, A, and I. To find the number of arrangements of these vowels, we treat them as indistinguishable because we cannot change their relative order. This can be calculated using the concept of permutations of indistinguishable items.

There are 4 vowels, so the number of arrangements is given by 4!.

2. Arrangement of Consonants:
The word EQUATION has 3 consonants - Q, T, and N. Similar to the arrangement of vowels, we treat these consonants as indistinguishable and calculate the number of arrangements using permutations of indistinguishable items.

There are 3 consonants, so the number of arrangements is given by 3!.

3. Total Arrangements:
To find the total number of different words, we multiply the number of arrangements of vowels by the number of arrangements of consonants since the arrangements of vowels and consonants are independent of each other.

Total arrangements = (number of vowel arrangements) * (number of consonant arrangements)
= 4! * 3!

Calculating the values,
4! = 4 x 3 x 2 x 1 = 24
3! = 3 x 2 x 1 = 6

Total arrangements = 24 * 6
= 144

Therefore, the number of different words that can be formed with the letters of the word EQUATION without changing the relative order of the vowels and consonants is 144.

However, none of the given options match with the correct answer of 144. Therefore, none of the provided options is correct for this question.
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Community Answer
How many different words can be formed with the letters of the word EQ...
The word "EQUATION" consists of 8 letters: E, Q, U, A, T, I, O, and N.
To find the number of different words that can be formed without changing the relative order of the vowels (EUAIO) and consonants (QTN), we need to calculate the number of arrangements for each group and then multiply them together.
The vowels (EUAIO) can be arranged among themselves, which gives us 5! = 5 x 4 x 3 x 2 x 1 = 120 arrangements.
The consonants (QTN) can be arranged among themselves, which gives us 3! = 3 x 2 x 1 = 6 arrangements.
Therefore, the total number of different words that can be formed is 120 x 6 = 720.
So, the correct answer is option C: 720.
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This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts: i. First, we will find out all the seven lettered words from the letters of word CLASSIC ii. Next, we will find out how many of these words will have the two Cs together. The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is: = 7 x 6 x 5 x 4 x ... x 1 = 7! Notice that there are two Cs and two S in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2! x 2! So, the total number of words formed is: 7!/(2! x 2!) We need to find how many of these words will have the two Cs together. To do this, let us treat the two Cs as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S which can be done by dividing 6! by 2!. Total number of 7 lettered words such that the two Cs are always together = 6!/2! The fraction of seven lettered words such that the two Cs are always together is: = (number of words with two Cs together/total number of words) = (6!/2!)/(7!/[2! x 2!]) = (2!/7) = 2/7 Hence the correct answer choice is A.

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How many different words can be formed with the letters of the word EQUATION without changing the relative order of the vowels and consonants?a)125b)620c)720d)880e)None of theseCorrect answer is option 'C'. Can you explain this answer?
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