The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

So, Required number of ways = \(\frac{6!}{(1!)(2!)(1!)(1!)(1!)}=360.\)

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 x ⁴C₃) + (⁶C₂ x ⁴C₂) + (⁶C₃ x ⁴C₁) + (⁶C₄)

= (⁶C₁ x ⁴C₁) + (⁶C₂ x ⁴C₂) + (⁶C₃ x ⁴C₁) + (⁶C₂)

=  \(\left(6\times4\right)+\left(\frac{6\times5}{2\times1}\right)+\left(\frac{6\times5\times4}{3\times2\times1}\right)+\left(\frac{4\times6}{2\times1}\times4\right) +\left(\frac{6\times5}{2\times1}\right)\)

= (24 + 90 + 80 + 15)

= 209.

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

So, Required number of numbers = (1 x 5 x 4) = 20.

Required number of ways = (⁸C₅ x ¹⁰C₆)

= (⁸C₃ x ¹⁰C₄)

\(\left(\frac{8\times7\times6}{3\times2\times1}\times\frac{10\times9\times8\times7}{4\times3\times2\times1}\right)\)

= 11760.

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

= (³C₁ x ⁶C₂) + (³C₂ x ⁶C₁) + (³C₃)

\(\left(3\times\frac{6\times5}{2\times1}\right)+\left(\frac{3\times2}{2\times1}\times6\right)+1\)

= (45 + 18 + 1)

= 64.

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.

To arrange the vowels (EAI) in the odd positions, we have 3 choices for the first odd position (either E or A or I) and 2 choices for the second odd position (one of the two remaining vowels).

Thus, we have 3 × 2 = 6 ways to arrange the vowels in the odd positions.

Now, we need to fill the even positions with the remaining consonants (D, T, and L). The consonants can be arranged in the remaining 3 even positions in 3! ways.

Therefore, the total number of ways to arrange the letters of the word DETAIL such that the vowels occupy only the odd positions is 6 × 3! = 36.

Hence, the correct answer is option C (36).

Required number of ways = (⁷C₅ x ³C₂) = (⁷C₂ x ³C₁) = \(\left(\frac{7\times6}{2\times1}\times3\right)= 63.\)

'LOGARITHMS' contains 10 different letters.

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

= ¹⁰P₄

= (10 x 9 x 8 x 7)

= 5040.

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.

So, Number of ways of arranging these letters = \(\frac{8!}{(2!)(2!)} = 10080.\)

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

Number of ways of arranging these letters = \(\frac{4!}{2!}= 12\)

So, Required number of words = (10080 x 12) = 120960.

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.

So,  Required number of ways = (120 x 6) = 720