Answer : Option D

Explanation:

 Video Explanation is given 

Answer : Option B

Explanation:

 Come from options

Answer : Option C

Explanation:

 As we discussed earlier question

The total number of square in a 8 x 8 chessboard is 204.

But in this case we have just counted 1 x 1 squares. What about the 2 x 2 squares, 3 x 3 squares, 4 x 4 squares and so on?

Number of 1 x 1 squares= 8 x 8 = 64 

Number of 2 x 2 squares =  7 x 7 = 49

Number of 3 x 3 squares =  6 x 6 = 36

Number of 4 x 4 squares =  5 x 5 = 25

Number of 5 x 5 squares =  4 x 4 = 16

Number of 6 x 6 squares =  3 x 3 = 9

Number of 7 x 7 squares =  2 x 2 = 4

Number of 8 x 8 squares =  1 x 1 = 1

Total number of Squares =  82 + 72 +62 +........... 22 + 12 = 204.

 Hence the probability 3 x 3 squares are selected = $\frac{64}{204} = \frac{16}{51}$

Answer : Option C

Explanation:

x + y + z = 10. Now, let us place 10 sticks in a row

|       |       |       |       |       |       |       |       |       |

This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance

|   +   |      |       |   +  |       |      |       |       |      |

This would be the equivalent of 1 + 3 + 6. or, x = 1, y = 3, z = 6.

There are 9 gaps between these 10 sticks, out of which one has to select 2 for placing the '+' symbol.

The number of ways of doing this would be 9C2. Bear in mind that this kind of calculation counts ordered triplets. (1, 3, 6) and (6, 3, 1) will both be counted as distinct possibilities. In general, we can also use this formula = sum-1Cn-1= 9C2 = 36.

Answer : Option A

Explanation:

Lets assume that P represents persons and empty boxes represents tasks.

Task 2 can be assigned in 2 ways (either to person 3 or person 4).

Now assume that ${\mathrm{P}}_3$ has occupied his place, then

Task 1 can then be assigned in only 3 ways (persons 4, 5 and 6) as ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$ cannot do the task 1

The remaining 4 tasks can be assigned to the remaining 4 persons in 4! = 24 ways

∴ The assignment can be done in 3 × 2 × 24 = 144 ways

Answer : Option B

Explanation:

In this question, given digits are 0 to 9, and the numbers has to be divisible by 3.

If the sum of the digits of a number is divisible by 3 then the number is divisible by 3.

Say: 12549 is divisible by 3, since sum of digits (1 + 2 + 5 + 4 + 9) = 21 / 3

In our question: 10 digits are available, with 10 places.

'0' cannot take the first position, as it is said 10 digit number.

Case 1: Starts with digit 1 and other digits can be ( 2 or 0 ) as sum = 3

 Case 2: Starts with digit 2 and other digits can be ( 1 or 0 ) as sum = 3

 Case 3: Starts with digit 3 and other digits must be (0) as sum = 3

Hence the total numbers formed are 9 + 36 + 9 + 1 = 55

Answer : Option C

Explanation:

First cube is painted with 4 green sides and 2 yellow sides.

Let the other cube have x sides with green and (6 - x) sides painted with yellow sides.

As per condition in the given question, when we roll both the dice, we can should have either green on both cubes or yellow on both cubes.

 4 green          2 yellow

X green         (6 – X) yellow

Both green or both yellow

$(\frac{4}{6} \mathrm{x} \frac{\mathrm{X}}{6}) +(\frac{{\displaystyle 2}}{{\displaystyle 6}} \mathrm{x} \frac{{\displaystyle \mathrm{X}-6}}{{\displaystyle 6}}) = \frac{1}{3}$  ( which is already given in the question)

$\mathit{\left(}\frac{\mathit{4}\mathit{x}\mathit{ }\mathit{-}\mathit{ }\mathit{12}\mathit{ }\mathit{+}\mathit{ }\mathit{2}\mathit{x}}{\mathit{36}}\mathit{\right)}\mathit{ }\mathit{=}\mathit{ }\frac{\mathit{1}}{\mathit{3}}$

$\left(\frac{6\mathrm{x} - 12}{12}\right) = 1$

x = 0

So second cube should not have any green faces.

Answer : Option C

Explanation:

For divisible by 15, the number can be divided by 3 and 5.

Divisible by 5, it should end with 0 or 5

Divisible by 3, (Sum of the digits) / 3

Case 1: The number is ending with 5 and the remaining places 0 cannot occupy

           $\cancel{0}$,  1,  2,  3,  4,  5

            _1_    _0_    _5_

            _1_    _3_    _5_

            _3_    _1_    _5_

            _3_    _4_    _5_

            _4_    _0_    _5_

            _4_    _3_    _5_

Here the total 6 numbers can be formed. 

Case 2: The number is ending with 0

             0,  1,  2,  3,  4,  5

          _1_    _2_    _0_

            _1_    _5_    _0_

            _2_    _1_    _0_

            _2_    _4_    _0_

            _4_    _2_    _0_

            _4_    _5_    _0_

            _5_    _1_    _0_

            _5_    _4_    _0_

Here, 8 numbers can be formed.

Total = 6 + 8 = 14 numbers

Answer : Option C

Explanation:

Given 11 alphabets A, H, I, M, O, T, U, V, W, X, Y.

4 out of the 11 symmetrical alphabets need to be selected,

Also the order of alphabets needs to be taken into account (as it is a password)

∴ 11P4 = 7920.

Hence, option 3.