How many 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 of a together?

In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.

Since all the vowels and consonants have to occur together, both [AEIOU] and [QTN] can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be `""^2P_2 = 2!`

Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words = 2! × 5! × 3!

= 1440

Solution

There are 8 letters in the word EQUATION including 5 vowels and 3 consonants.

Now 5 vowels can be arranged in 5! ways and 3 consonants can be arranged in 3! ways. Also the two groups of vowels and consonants can be arranged in 2! ways.

∴ Total number of permutations

= 5!×3!×2!=120×6×2=1440.

Solution : There are 5 vowels and 3 consonants in the word 'EQUATION'. If all 5 vowels occur together and all 3 consonants occur together then taking them as one-one letter, no. of ways to arrange them
`= .^[2]P_[2] =2! = 2`
Again, no of ways of arranging 5 vowels `= 5! = 120`
No. of ways of arranging 3 consonants `= 3! = 6`
Therefore, total no. of words `= 2 xx 120 xx 6 = 1440`.

Misc 2 - Chapter 7 Class 11 Permutations and Combinations [Term 2]

Last updated at Jan. 30, 2020 by

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Misc 2 How many 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? Misc 2 How many 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? Number of vowels in EQUATION = E, U, A, I, O = 5 Number of ways vowels can be arranged = 5P5 = 5!/[5 − 5]! = 5!/0! = 5!/1 = 120 Number of consonants in EQUATION = Q, T, N = 3 Number of ways consonants can be arranged = 3P3 = 3!/[3 − 3]! = 3!/0! = 3!/1 = 6 Total number of ways in which vowels & consonants occur together = 2 × [Number of ways vowel arrange] × [Number of ways consonants arrange] = 2 × [120 × 6] = 1440

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