How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2  -  6x5x4/3x2

= 84 - 20

= 64 ways.

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.

Given in the question that, there are 7 boys and 6 girls. 

Team members = 5

Now, required number of ways in which a team of 5 having atleast 3 girls in the team = 

$$\begin{gathered} 6C3 x 7C2 + 6C4 x 7C1 + 6C5 \\ = 6x5x43x2x1 x 7x62x1 + 6x5x4x34x3x2x1 x 7 + 6x5x4x3x25x4x3x2x1 \\ = 420 + 105 + 6 \\ = 531. \end{gathered}$$