Permutation-Combination
Aptitude

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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.

${^{12}C_4} × {^4C_3}$

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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,

$= {^{12}C_4} × {^4C_3} × 7!$


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