How many words with or without meaning can be formed using all the letters of the word utopian at a time so that the vowels and consonants occur 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

Misc 2 - Chapter 7 Class 11 Permutations and Combinations (Term 2)

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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

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?

Answer

Verified

Hint: Permutations are the different ways in which a collection of items can be arranged. For example:
The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC. Note that ABC and CBA are not the same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.
The number of ways in which n things can be arranged, taken all at a time, \[{}^n{P_n}{\text{ }} = {\text{ }}n!\], called ‘n factorial.’

Complete step-by-step answer:
Total number of letters in “EQUATION” = 8.
There are 5 vowels: a, e, i, o, u and 3 consonants : q, t, n.
Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects.
Then they form 2 groups V(vowels) and C (consonants)
We first arrange the 2 groups.
The permutations of these 2 objects taken all at a time are counted: ${}^2{P_2} = 2! = 2$ways
Corresponding to each of these permutations,
Now the group V has 5 elements, they can be arranged in $5! = 120$ ways.
Now the group C has 3 elements, they can be arranged in $3! = 6$ ways.
Hence by multiplication principle, required number of words = $2! \times 5! \times 3!$
the total no of ways = $1440$

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.

Note: Always keep an eye on the keywords used in the question. The keywords can help you get the answer easily.
The keywords like-selection, choose, pick, and combination-indicates that it is a combination question.
Keywords like-arrangement, ordered, unique- indicates that it is a permutation question.
If keywords are not given, then visualize the scenario presented in the question and then think in terms of combination and arrangement.

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