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)
	= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5 2 x 1 2 x 1 3 x 2 x 1 2 x 1
	= (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.

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

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

Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
	= 3 x 6 x 5 + 3 x 2 x 6 + 1 2 x 1 2 x 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.