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We make an assumption that any combination of letters is a word and we should take repetition into account because that would mean the same word.

How many words can be formed using all the letters in the word EXAMINATION in such a way that the first 2 are different consonants and the last 2 are different vowels.

Okay so here's my approach.

There are 4 different consonants and 6 vowels (with 2 Identical I's and A's)

So ${{4} \choose {2}}(2!)$ would be the consonants and ${6 \choose 2}(2!) $would be the vowels (still including repetition for now)

Now I'd take these two and times by 7! All over 2!2!2! To account for repetition and get 226800. Is this correct?

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  • $\begingroup$ Try some more. Why ${6\choose 2}$ for different vowels ? What if the starting consonants are NX ? $\endgroup$ – true blue anil Jun 11 '15 at 13:41
  • $\begingroup$ Don't forget that there is also a duplicate consonant in N. $\endgroup$ – hypergeometric Jun 11 '15 at 14:48
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(Having thought through further, this is a shorter solution, and in some ways similar to that posted by true blue anil earlier; the original answer posted is given after this one)

  • Assume all letters are different, i.e. no repeats. e.g. the two $N$s are $N_1$ and $N_2$.
  • Head: Number of arrangements with consonants only =$^5P_2$.
  • Removing arrangements for same base consonants ($N_1 N_2, N_2 N_1$) gives $^5P_2-2$.
  • Body: Number of arrangements =$7!$.
  • Tail: Number of arrangements with vowels only =$^6P_2$.
  • Removing arrangements for same base vowels ($I_1 I_2, I_2I_1, A_1A_2, A_2A_1$) gives $^6P_2-4$.
  • Overall: Total number of arrangements, normalising for repeats, is given by

$$\frac {({^5P_2}-2)\cdot 7!\cdot ({^6P_2}-4)}{2!\ 2!\ 2!}=\frac {18\cdot 7!\cdot 26}{8}=294840\;\blacksquare$$


(What follows below is the original answer posted, which is a bit longer and uses a different approach)

We note the following points:

  • Because of the way the question is structured, we can focus on the number of ways to construct the Head (2 letters), the Tail (2 letters) and the Body (7 letters).
  • There are no repeats in the Head and Tail, so repeats occur - if they do - only in the Body.

Head

  • (a) If the Head contains 1 N ($3\times 2=6$ arrangements), then there is $0$ repeat N in the Body
  • (b) If the Head contains 0 N ($^3P_2=6$ arrangements), then there is $1$ repeat N in the Body.

Tail

  • (1) If the Tail contains one 1 A and 1 I ($2$ arrangements: AI, IA), then there is $0$ repeat A or I in the Body

  • (2) If the Tail contains either 1 A or 1 I ($8$ arrangements: $^4P_3-2-2$), then there is $1$ repeat I or A respectively in the Body

  • (3) If the Tail contains 0 A and 0 I ($2$ arrangements: EO, OE), then there are $2$ repeats (one each for A and I) in the Body.

Body

  • The number of arrangements of letters in the body depends on the number of repeats and is given by $7!/(2!)^n$ where $n$ is the number of repeats in the body for a given Head-Tail combination, contributed by both the Head and the Tail.

Permutations

The number of arrangements or permutations given by different Head-Tail combinations are as follows:

(a) (1): $6\times 2\times {7!}/{(2!)^{0+0}}\color{lightgrey}{=6\times 16\times 7!/(2!)^3}=\;\;60480$
(a) (2): $6\times 8\times {7!}/{(2!)^{0+1}}\color{lightgrey}{=6\times 32\times 7!/(2!)^3}=120960$
(a) (3): $6\times 2\times {7!}/{(2!)^{0+2}}\color{lightgrey}{=6\times \ \ 4 \times 7!/(2!)^3}=\;\;15120$
(b) (1): $6\times 2\times {7!}/{(2!)^{1+0}}\color{lightgrey}{=6\times \ \ 8\times 7!/(2!)^3}=\;\;30240$
(b) (2): $6\times 8\times {7!}/{(2!)^{1+1}}\color{lightgrey}{=6\times 16\times 7!/(2!)^3}=\;\;60480$
(b) (3): $6\times 2\times {7!}/{(2!)^{1+2}}\color{lightgrey}{=6\times \ \ 2 \times 7!/(2!)^3}=\;\;\;\;7560$

Total number of arrangements $\color{lightgrey}{=6\times 78\times 7!/(2!)^3} =294840\;\; \blacksquare$

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  1. At the 2 ends, for both consonants and vowels,

    there will be $^4P_2$ = 12 patterns, producing 12*12 = 144 patterns

  2. N is 1 of 4 distinct consonants, so in 3 cases correction for duplication will not be needed.

  3. A & I are 2 distinct vowels in 4, so in 6 cases, correction for one duplication will not be needed.

  4. So full correction for duplication by division by 2!2!2! will be needed in 9*6 = 54 cases, a duplication correction for consonant will be cancelled in 3*6 = 18 cases, for a vowel in 9*6 = 54 cases and for both vowel and consonant in 3*6 =18 cases

  5. We can weight the 144 cases to get

    ans = 7!($\frac{54}{8} + \frac{18}{4} + \frac{54}{4} +\frac{18}{2}$) = 170,100

Edit

For some obscure reason, I took the extreme consonant and vowel as being outside the purview of the duplication, so the answer above is incorrect.

Edit 2

Correcting for the misinterpretation, let us make duplicate letters in lower case to start with, thus X M T N n for consonants and E A a I i O for vowels. So $^5C_2$ -1 = 9 combos at one end & $^6C_2$ - 2 = 13 combos at the other, excluding ones where Nn, Aa or Ii are at the ends. Arrangements at the ends will be 18 & 26 respectively.

Thus 18*26*7!/(2!)³ = 294,840

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  • 1
    $\begingroup$ Looks like we arrived at the same answer! $\endgroup$ – hypergeometric Jun 12 '15 at 16:19
  • $\begingroup$ Good ! Initially I was also basically using the head - body - tail approach, but got a bit confused so switched to another one. $\endgroup$ – true blue anil Jun 12 '15 at 17:57

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