Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?
 ✖ A. ${^{16}C_7} × 7!$ ✔ B. ${^{12}C_4}×{^4C_3}×7!$ ✖ C. ${^{12}C_3} × {^4C_4}$ ✖ D. ${^{12}C_4} × {^4C_3}$

Solution:
Option(B) is correct

4 consonants out of 12 can be selected in ${^{12}C_4}$ ways.

3 vowels can be selected in ${^4C_3}$ ways.

Therefore, total number of groups each containing 4 consonants and 3 vowels,

$= {^{12}C_4} × {^4C_3}$

Each group contains 7 letters, which can be arranging in $7!$ ways.

Therefore, required number of words,

$= {^{12}C_4} × {^4C_3} × 7!$