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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

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?

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1 Answer

up vote 2 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
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