Select any four points in 6C4 = 15 ways .

All distinct a < b < c . 10C3 => 120 two same 10C1 x 9C1 => 90 cases a=b = c 10 cases Total : 120 + 90 + 10 = 220.

Three cases: (1) When two fours go to black boxes. for 4's => 1 way , for 1 , 2, and 3 there are 3! ways . (2) When one four goes to one black box...

The probability of 0 hits out of n is nCo × (1/2)^n and 1 hit out of n is nC1 × ( 1/2)^n Probability of destroying > 0.99...... = 1-{nC0×(1/2)^n + ...

15.is it 33 ?

14. E is four steps ahead of A. So it is not possible to reach E from A in (2n - 1) i.e. odd number of steps. Hence Option ( A)

PFA the solution

AD ( Altitude of the triangle ) = AB sqrt{3}/2 AD = 16sqrt{3}/2 = 8sqrt 3AE = ED = 4 sqrt{3} BE = sqrt { DE^2 + BD^2 } = sqrt {[ 8^2 + 4(sqrt3)^2 ]

Use this : CP^2 = AP x PB and EQ^2 = AQ x QB .

Number of tiles = Area of the floor / ( Area of a tile )

PFA the Approach :

Alternative approach : AD^2 = BD x DC 4^2 = 6 x DC DC = 16/6 = 8/3 AC = AD + DC = 8/3 + 6 = 26/3