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