How many words each containing 3 vowels and 2 consonants can be formed from the letters of the word involute?

How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word ‘INVOLUTE’?

How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word ‘INVOLUTE’?

How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word ‘INVOLUTE’?

Answer : In the word ‘INVOLUTE’ there are 4 vowels, ‘I’,’O’,’U’ and ‘E’ and there are 4 consonants, ‘N’,’V’,’L’ and ‘T’. 3 vowels out of 4 vowels can be chosen in 4C3ways. 2 consonants out of 4 consonants can be chosen in 4C2 ways. Length of the formed words will be (3 + 2) = 5. So, the 5 letters can be written in 5! Ways. Therefore, the total number of words can be formed is = (4C3 X 4C2 X 5!) = 2880.

Given as

The word ‘INVOLUTE’

The total number of letters = 8

The total vowels are = I, O, U, E

The total consonants = N, V, L, T

Therefore number of ways to select 3 vowels is 4C3

And the number of ways to select 2 consonants is 4C2

Then, the number of ways to arrange these 5 letters = 4C3 × 4C2 × 5!

By using the formula,

nCr = n!/r!(n – r)!

4C3 = 4!/3!(4 - 3)!

= 4!/(3! 1!)

= [4 × 3!] / 3!

= 4

4C2 = 4!/2!(4 - 2)!

= 4!/(2! 2!)

= [4 × 3 × 2!] / (2! 2!)

= [4 × 3] / (2 × 1)

= 2 × 3

= 6

Therefore, by substituting the values we get

4C3 × 4C2 × 5! = 4 × 6 × 5!

= 4 × 6 × (5 × 4 × 3 × 2 × 1)

= 2880

∴ The no. of words that can be formed containing 3 vowels and 2 consonants chosen from ‘INVOLUTE’ is 2880.

Given,

The word is INVOLUTE. 

We have 4 vowels namely I,O,U,E, and consonants namely N,V,L,T. 

We need to find the no. of words that can be formed using 3 vowels and 2 consonants which were chosen from the letters of involute. 

Let us find the no. of ways of choosing 3 vowels and 2 consonants and assume it to be N1. 

⇒ N1 = (No. of ways of choosing 3 vowels from 4 vowels) × (No. of ways of choosing 2 consonants from 4 consonants) 

⇒ N1 = (4C3) × (4C2) 

We know that,

nCr = \(\frac{n!}{(n-r)!r!}\)

And also,

n! = (n)(n – 1)......2.1

⇒ N1 = 4 × 6 

⇒ N1 = 24

Now,

We need to find the no. of words that can be formed by 3 vowels and 2 consonants. 

Now,

We need to arrange the chosen 5 letters. 

Since every letter differs from other. 

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 5!. 

Let us the total no. of words formed be N. 

⇒ N = N1 × 5! 

⇒ N = 24 × 120 

⇒ N = 2880 

∴ The no. of words that can be formed containing 3 vowels and 2 consonants chosen from INVOLUTE is 2880.

Solution : There are 4 vowels namely, I, O, E, U
and 4 consonants, namely N, V, L, T
`therefore` 3 out of 4 vowels an be selected in `.^(4)C_(3)=4` ways
2 out of 4 consonants can be selected in `.^(4)C_(2)=6` ways
`therefore` Number of combinations of 3 vowels and 2 consonants `= 4xx6=24`
Now in each of these 24 combinations, 5 letters can be arranged among themselves in 5! ways
= 120 ways.
`therefore` Required number of different words `= 24xx120 =2880`.

Solution : vowels= I,O,U,E
consonants = N, V,L,T
total words`= (.^4C_3 .^4C_2) .^5P_5`
`= (4!)/(1!*3!)*(4!)/(2!*2!)*(5!)/(5-5)!`
`= 4 * (4*3*2)/4 * 5!`
`= 24*120`
`=2880`
answer

Example 20 - Chapter 7 Class 11 Permutations and Combinations (Term 2)

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Example 20 How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ? Thus, Number ways of selecting 3 vowels & 2 consonants = 4C3 × 4C2 = 4!/3!1! × 4!/2!2! = (4 × 3!)/(3! × 1!) × (4 × 3 × 2 × 1)/(2 × 1 × 2 × 1) = 4 × 6 = 24 We have selected the letters, Now, we have to arrange Number of arrangements of 5 letters Number of arrangements of 5 letters = 5P5 = 5!/(5 − 5)! = 5!/0! = 5!/1 = 5 × 4 × 3 × 2 × 1 = 120 Thus, Total number of words = Number of ways of selecting × Number of arrangements = 24 × 120 = 2880

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