Solution:
In the given word EQUATION,
- No. of vowels = 5 (E, U, A, I, O)
- No. of consonants = 3 (Q, T, N)
Let us consider that there are 2 units, one consisting of vowels and one of the consonants. i.e.,
No. of ways in which 5 vowels can be arranged among themselves = 5!
No. of ways in which 3 consonants can be arranged among themselves = 3!
No. of ways in which these two units can be arranged = 2!
The total number of ways = 2! × 3! × 5! = 1440
NCERT Solutions Class 11 Maths Chapter 7 Exercise ME Question 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?
Summary:
The number of words, with or without meaning, that can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together is 1440
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 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 <br> `= .^(2)P_(2) =2! = 2` <br> Again, no of ways of arranging 5 vowels `= 5! = 120` <br> No. of ways of arranging 3 consonants `= 3! = 6` <br> Therefore, total no. of words `= 2 xx 120 xx 6 = 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.
Misc 2 - Chapter 7 Class 11 Permutations and Combinations (Term 2)
Last updated at Jan. 30, 2020 by
This video is only available for Teachoo black users
Solve all your doubts with Teachoo Black (new monthly pack available now!)
Transcript
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