How many words can be made from the letters M N C F A E I O U such that a group of exactly two vowels is together? How many words can be made from the letters M N C F A E I O U such that a group of exactly two vowels is together?
I tried to solve it as : 
no. of ways in which we can arrange $4$ consonants is $4!$, so there are $5$ places for $4$ vowels but there can be one or two group of vowels so 
when we have one pair of vowels no. of words is $4! \cdot 5p3 \cdot 4c2  \cdot 2$
when we have two pair of vowels no.of words is $4! \cdot 5p2 \cdot 4c2 \cdot 2c2 \cdot 2 \cdot 2$
 A: Since there are five vowels, we can have one or two pairs with exactly two vowels together.  If we have one group with exactly two vowels together, then the vowels can be distributed in two ways, a group of two and a group of three or as a group of two and three single vowels.  If we have exactly two groups with exactly two vowels together, then the vowels must be distributed as two groups of two and a single vowel.
A group with exactly two vowels and a group with exactly three vowels:  There are $4!$ ways of arranging the consonants, which creates five spaces in which to place the vowels, three between successive consonants and two at the ends of the row. There are five ways to choose the space in which exactly two vowels will be placed and four ways to choose one of the remaining four spaces in which to place three vowels.  The vowels can then be arranged from left to right in the selected spaces in $5!$ ways.  Hence, the number of such arrangements is 
$$4! \cdot \binom{5}{1} \cdot \binom{4}{1} \cdot 5!$$
A group with exactly two vowels and three single vowels:  Again, there are $4!$ ways to arrange the consonants, which creates five spaces.  Thus, there are five ways to choose the space in which exactly two vowels will be placed.  There are $\binom{4}{3}$ ways to choose three of the four remaining spaces for the single vowels.  The vowels can then be arranged from left to right in the selected spaces in $5!$ ways.  Hence, the number of such arrangements is 
$$4! \cdot \binom{5}{1} \cdot \binom{4}{3} \cdot 5!$$
Two groups with exactly two vowels and a single vowel: As before, there are $4!$ ways to arrange the consonants, which creates five spaces.  There are $\binom{5}{2}$ to choose two of these spaces in which to place exactly two vowels and three ways to choose one of the remaining three spaces for the single vowel.  The vowels can then be arranged from left to right in the selected spaces in $5!$ ways.  Hence, the number of such arrangements is 
$$4! \cdot \binom{5}{2} \cdot \binom{3}{1} \cdot 5!$$
Since the three cases are mutually exclusive and exhaustive, the desired answer is found by adding the above cases.   
A: Hints.(If the question asks for at least one pair of vowels and doesn't specify the number of letters to be used in the formation of word)  
1)Number of two letter words:
You need to choose two vowels from 5 and arrange them.
Hence ways =$\binom {5}{2}.2!$
2) Number of three letter words:
Choose two vowels and one consonant and arrange them such that vowels are together. 
Hence ways= $\binom {5}{2}.2!.2!$
3) Number of 4 letter words:
Using same idea above you get ways=$\binom {5}{2}\binom {4}{2}.2!.3!$
4) Number of 5 letter words:
Case 1) Choose 4 vowels and 1 consonant and then arrange the pair of vowels on either side of consonant. 
Hence ways =$\binom {5}{4}\binom {4}{1}.\left( \frac {4!}{2!2!2!}\right). 2!$
Case 2) Choose two vowels and 3 consonants and arrange them accordingly. 
Hence ways=$\binom {5}{2}\binom {4}{3}.4!.2!$
You can continue for further cases if any. 
