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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


Excluding N ending in EA, EI, AO or IO


Excluding N ending in EO


Including N ending in AA or II:


Including N ending in AI:


Including N ending in EA, EI, AO or IO:


Including N ending in EO:


Summing all these cases up yields:

219 240

Am I overcounting?

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up vote 3 down vote accepted

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.

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So are you saying that the solution of 189 000 that I have is wrong? Now I am getting confused. Also where do 6 and 5 come from when you allocate vowels? Could you please put annotations in brackets next to your working so that at least I know what you are doing and why you are doing it. – Jquery Ninja Apr 29 '13 at 23:59
@Jquery: How do you mean "at least"? – joriki May 1 '13 at 9:22
you are dealing with less than an armateur, i.e. a beginner. To you 6.5.4C2 and 5.4.3 might seem obvious but I can't exactly put my finger on how you came up with those values. At least means if instead you edit your answer and write: 6 (choose the first vowel). 5(the second vowel according to the fundamental counting principle).4C2(choose any 2 out of the 4 remaining vowels) and maybe instead of 5.4.3 you could edit it to 5(place first consonant in the remaining 7 positions or is it the first fixed??).4(the second) – Jquery Ninja May 2 '13 at 2:50
@Jquery: I added some details -- hope that makes it clearer? – joriki May 5 '13 at 22:09
that solution was wrong indeed I will show you how I eventually did it the long way. It's close to your answer though – Jquery Ninja Jun 11 '13 at 15:31

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.

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