Question

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Answer 1

Hint: Count the number of vowels and consonants in the word DAUGHTER. Let the counts be x, y respectively. The required words should have 2 vowels and 3 consonants in it. So the no. of words that contains 2 vowels and 3 consonants which can be formed from the letters of DAUGHTER is ${}^x{C_2} \times {}^y{C_3}$

Complete step-by-step answer:
We are given to find the number of words that can be formed from the letters of the word DAUGHTER which contains 2 vowels and 3 consonants.
The given word is DAUGHTER. This word has 3 vowels, A, U, E, and 5 consonants, D, G, H, T and R.
So the required words should have 2 vowels from A, U and E; 3 consonants from D, G, H, T and R.
And the order of the letters is not specific, which means the letters can be used in any order. So we have to use combinations.
So the no. of words will be ${}^3{C_2} \times {}^5{C_3}$, selecting any 2 from 3 vowels and selecting any 3 from 5 consonants.
$
  {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\
  {}^3{C_2};n = 3,c = 2 \\
  {}^3{C_2} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1!}} = \dfrac{6}{2} = 3 \\
  {}^5{C_3};n = 5,c = 2 \\
  {}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2!}} = \dfrac{{120}}{{12}} = 10 \\
  \therefore No.of words = {}^3{C_2} \times {}^5{C_3} = 3 \times 10 = 30 \\
$
Therefore, 30 words can be formed from the letters of the word DAUGHTER each containing 2 vowels and 3 consonants.

Note: A Permutation is arranging the objects in order. Combinations are the way of selecting the objects from a group of objects or collection. When the order of the objects does not matter then it should be considered as Combination and when the order matters then it should be considered as Permutation. Do not confuse using a combination, when required, instead of a permutation and vice-versa.