Find the value of 'n' for which the nth term of two AP'S:

15,12,9.... and -15,-13,-11...... are equal?

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.

Required number of ways = 7C5 × 3C2=63

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.

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways=3C1*6C2+3C2*6C1+3C3 = (45 + 18 + 1) =64

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

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). 

Required number of ways = 6C1*4C3+6C2*4C2+6C3*4C1+6C4  

= 6C1*4C1+6C2*4C2+6C3*4C1+6C2 = 209.

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.