Counting Methods: Restricted Permutations I have been scratching my head for a long time. The question is: How many words can be formed using all letters in the word EXAMINATION in such a way that the first two letters are different consonants while the last two letters are vowels?
Solution: 189 000
My attempt:
Consonants:  XMNNT
Vowels: EOAAII
For the vowels there is no restriction and there are 8 possible combinations:
AA
II
AI
EA
EI
EO
AO
IO
For the consonants there has to be no repetition so there are two cases, one including N and one excluding N.
Excluding N ending in AA or II:
3 * 2 * 7! * 1 * 1 * 2 / 2!
Excluding N ending in AI
3*2*7!*1*1/2!
Excluding N ending in EA, EI, AO or IO
3*2*7!*2*2/2!
Excluding N ending in EO
3*2*7!*1*1/2!^3
Including N ending in AA or II:
1*3*7!*1*1*2
Including N ending in AI:
1*3*7!*1*1
Including N ending in EA, EI, AO or IO:
1*3*7!*1*1*4
Including N ending in EO:
1*3*7!*1*1/2!^2
Summing all these cases up yields:
219 240
Am I overcounting?
 A: There are $6$ vowels and $5$ consonants. The positions of $2$ vowels and $2$ consonants are fixed, so there are $\binom73=35$ ways to choose the $3$ remaining consonant positions from the $7$ remaining positions. There are no further restrictions on the vowels, so there are $6\cdot5\cdot\binom42=180$ different ways to arrange the vowels on the vowel positions. Without restrictions, there would be $5\cdot4\cdot3=60$ ways to arrange the consonants on the consonant positions, but we have to subtract the $3!=6$ arrangements in which the two $N$s are at the beginning, leaving $60-6=54$ options. Thus the total number of admissible arrangements is $35\cdot180\cdot54=340200$.
[Edit in response to the comments:]
There are $11$ letters in total. The first two letters are consonants and the last two letters are vowels. That leaves $3$ consonants in the $7$ remaining positions, and there are $\binom73$ ways to choose $3$ positions out of $7$.
Since the vowels come in two singletons and two pairs, after $6$ positions have been selected for the $6$ vowels there are $6$ possible positions for the first singleton, $5$ for the second singleton and then $\binom42$ for one of the pairs (leaving no further choice for the positions of the remaining pair), for a total of $6\cdot5\cdot\binom42$ positioning options for the vowels. The consonants come in three singletons and one pair, so after $5$ positions have been selected for the $5$ consonants there are $5$ possible positions for the first singleton, $4$ for the second and $3$ for the third (leaving no further choice for the positions of the pair), for a total of $5\cdot4\cdot3$ positioning options for the consonants.
A: just seen this 'old' post and here is my solution: 

The list of the letters must be (2 consonants 7 letters and 2 vowels): 
(0)  cc  lllllll  vv  

(1)  cc has no 'N', there are 6 possibilities: XT XM MT MX TM TX  
(2)  cc has a 'N',  there are 7 possibilities: NN NT NM NX XN MN TN   
(3)  vv has 'A' and' 'I', 2 possibilities: AI IA   
(4)  vv has 'A' but no 'I', 5 possibilities: AA AO AE EA OA  
(5)  vv has 'I' but no 'A', 5 possibilities: II IO IE EI OI  
(6)  vv has no 'A' nor 'I', 2 possibilities: EO OE  
(7)  lllllll with all different letters, 7! possibilities 
(8)  lllllxx with 2 identical letters, 7!/2 possibilities 
(9)  lllxxyy (2 x 2 indentical letters), 7!/4 possibilities 
(10) lxxyyzz (3 x 2 identical letters), 7!/8 possibilities 

Now the 3 fields in the sheme (0) are:  
Nc lllllll AI   (2) (7) (3): 7.7!.2 = 14.7! possibilities  
 Nc lllllII Av   (2) (8) (4): 7.(7!/2).5 = (35/2).7! possibilities  
 Nc lllllAA Iv   (2) (8) (4): 7.(7!/2).5 = (35/2).7! possibilities  
 Nc lllIIAA vv   (2) (9) (6): 7.(7!/4).2 = (7/2).7! possibilities  
cc lllllNN AI   (1) (8) (3): 6.(7!/2).2 = 6.7! possibilities  
 cc lllNNII Av   (1) (9) (4): 6.(7!/4).5 = (15/2).7! possibilities 
 cc lllNNAA Iv   (1) (9) (4): 6.(7!/4).5 = (15/2).7! possibilities 
 cc lNNIIAA vv   (1) (10) (6): 6.(7!/8).2 = (3/2).7! possibilities 
Adding these 8 configurations, we get 75.7! = 378000 possibilities 
BTW, 378000 = 2.189000 , probably a forgotten permutation in the given solution. 
Cheers. 
