The word 'OPTICAL' contains 7 different letters.

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

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

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

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

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.

Required number of ways= 8C5×10C6 = (8C3×10C4)= 11760.

The PS command prints the process status for some or all of the running processes

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.