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?
Best Answer
(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)
$$\frac {({^5P_2}-2)\cdot 7!\cdot ({^6P_2}-4)}{2!\ 2!\ 2!}=\frac {18\cdot 7!\cdot 26}{8}=294840\;\blacksquare$$
(What follows below is the original answer posted, which is a bit longer and uses a different approach)
We note the following points:
Head
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
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 $\color{lightgrey}{=6\times 78\times 7!/(2!)^3} =294840\;\; \blacksquare$