Question

Five letter words are formed each containing $2$ consonants and $3$ vowels out of the letters of the word EQUATION. In how many of these the two consonants are always together?

Answer 1

Total $3C$ and $5V$
Choose $2C$ and $3V$
Total ${}^3{C}_2\cdot {}^5{C}_3\cdot 5!=3.10\times 120=3600$
Two consonants together. Tie them after selecting 2 consonants and then untie.
There will be 3 vowels and one unit of third consonant. In all four units which can be arranged in $4!$ ways. Hence the total number of words will be
${}^3{C}_2\cdot {}^5{C}_3\cdot 4!\times 2!=3\times 10\times 24\times 2=1440$
Alternative method:
Total-Never together
3 vowels can be arranged in ${}^5{P}_3=60$ ways.
2 consonants can be selected in ${}^3{C}_2=3$ ways.
There will be 4 gaps in between the vowels in which two can be arranged in ${}^4{P}_2=4.3=12$.
Hence the total when consonants are separated
$=60\times 3\times 12=2160$.
Required number when the consonants are together
$=3600-2160=1440$