In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =4!/2!= 12.

Required number of words = (10080 x 12) = 120960

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= 5040.

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under: 

                                                      (1) (2) (3) (4) (5) (6) 

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.  

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions. 

Number of ways of these arrangements = 3P3 = 3! = 6. 

Total number of ways = (6 x 6) = 36.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3*4C2) 

= 210. 

Number of groups, each having 3 consonants and 2 vowels = 210. 

Each group contains 5 letters. 

Number of ways of arranging 5 letters among themselves = 5! = 120 

Required number of ways = (210 x 120) = 25200.

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =7!/2!= 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 3!/5!= 20 ways.

Required number of ways = (2520 x 20) = 50400.

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

216004989600=0.043

3 letters can be choosen out of 9 letters in 9C3 ways.

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in (4C2*5C1)+4C3 ways

Hence,required probability = 4C2*5C1+4C39C3 = 17/42