Answer : Option D

Explanation:

 In a 8 x 8 chessboard ( grid ) given below

We have 1 x 1 squares, 2 x 2 squares,3 x 3 squares,4 x 4 squares ......................... 8 x 8 squares.

So, the number of total square in a N x N grid is 

$1^2 + 2^2 + 3^2 + 4^2 ............ {\mathrm{N}}^2$.

Hence for the 8 x 8 chessboard, total type of squares 

$1^2 + 2^2 + 3^2 + 4^2 ............ 8^2$.

1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204

Answer : Option C

Explanation:

This can be done by using a formula:

The number of rectangles formed in a m x n grid is given by

( 1 + 2 + 3 + 4 ...... m ) x ( 1 + 2 + 3 + 4 ...... n ), where m & n are the size of the grid.

In our question, it is given the 5 x 6 grid means

m = 5 and n = 6

So, the number of rectangles

$\Rightarrow$( 1 + 2 + 3 + 4 + 5 ) x ( 1 + 2 + 3 + 4 + 5 + 6 ) 

$\Rightarrow$( 15 ) x ( 21 )  

$\Rightarrow$315 rectangles are formed 

Answer : Option C

Explanation:

As we discussed earlier question[1]

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 =  $8^2 + 7^2 +6^2 +........... 2^2 + 1^2$ = 204.

 Hence the probability 3 x 3 squares are selected = $\frac{36}{204}$ = $\frac{3}{17}$

Answer : Option D

Explanation:

The white square can be chosen in 32 ways.

If we remove the row and the column which contains the chosen white square, we will be left with 7 Rows & 7 Columns containings a total of 49 squares (24 black, 25 whites)

Say, initially we choose White square (W1), then from the 7 x 7 remaining square, we have to select any black square ( 24 possibilities)

So each W1 x 24 black squares,

therefore, each white square x 24 possibilities

Required ways = 32 x 24 = 768

Answer : Option A

Explanation:

If Neelam wants to reach B in the shortest possible path, then she has to travel in the diagonal path say ( X $\to$Y)

(i.e) ( A $\to$X) ( X $\to$Y) and finally ( Y $\to$B)      

Below diagram shows the partially routes

Let us take ( A$\to$X), here we can see the grid is ( 2 x 2 )

Number of ways = $\frac{\mathbf{(}\mathbf{m}\mathbf{+}\mathbf{n}\mathbf{)}\mathbf{!}}{\mathbf{m}\mathbf{!}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{n}\mathbf{!}}$ = $\frac{\mathbf{(}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{!}}{\mathbf{2}\mathbf{!}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{2}\mathbf{!}}$ = 6 ways

Let us take ( X$\to$Y), here we have one straight diagonal path

Number of ways = 1 way.

Let us take ( B$\to$Y), here we can see the grid is ( 2 x 4 )

Number of ways = $\frac{\mathbf{(}\mathbf{m}\mathbf{+}\mathbf{n}\mathbf{)}\mathbf{!}}{\mathbf{m}\mathbf{!}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{n}\mathbf{!}}$= $\frac{\mathbf{(}\mathbf{2}\mathbf{+}\mathbf{4}\mathbf{)}\mathbf{!}}{\mathbf{2}\mathbf{!}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{4}\mathbf{!}}$ = 15 ways

Therefore, the total number of paths possible = 6 x 1 x 15 = 90 ways.

Answer : Option B

Explanation:

For any given number to be divisible by 8, its last three digits must divisible by 8.

So, last three digits can be from the following condition:

(024), (048), (064), (208), (240), (248), (264), (280), (408), (480), (608), (624), (640), (648), (680), (824), (840), (864),

There are total 18 such cases.

Now, Consider these 12 cases in which zero is there

(024), (048), (064), (208), (240), (280), (408), (480), (608), (640), (680), (840)

Lets the number ends with  0 2 4

With each of these 12 cases, first digit can be any of the remaining 2 digits and second digit can be the remaining one digit

Therefore, 12 × 2! = 24 such numbers

Consider the 6 cases in which zero does not come.

(248), (264), (624), (648), (824), (864)

 Lets the number ends with  2 4 8       

With each of these 6 groups, first digit can be the one remaining digit(because zero cannot be the first digit)

Second digit can be zero

 Therefore, 6 × 1 = 6 such numbers

Hence total numbers formed 24 + 6 = 30

Answer : Option C

Explanation:

The explanation is given in the Video 

Answer : Option A

Explanation:

In general, the sum of the digits is given by 

(n - 1)! x (111...1 n times) x (sum of digits), where n is number of digits

So the given digits are 1, 2, 3, 4 and 5

$\to$(5 - 1)! x (11111) x (1+2+3+4+5)

$\to$(4)! x (11111) x (15)

$\to$(24) x (11111) x (15)

$\to$3999960

Answer : Option B

Explanation:

 The total 50 paise can be paid using these coins:

With any one type coins

50(1 paise)
10(5 paise)
5(10 paise)
2(25 paise)
total 4  ways -----(1)

With any two type coins

5(1 paise),  9(5 paise)
10(1 paise), 8(5 paise)
15(1 paise), 7(5 paise)
...
45(1 paise), 1(5 paise)       (9 ways)

10(1 paise),  4(10 paise)
20(1 paise), 3(10 paise)
30(1 paise), 2(10 paise)
40(1 paise), 1(10 paise)     (4 ways)

25(1 paise),  1(25 paise)    (1 ways)

2(5 paise),  4(10 paise)
4(5 paise),  3(10 paise)
6(5 paise),  2(10 paise)
8(5 paise),  1(10 paise)      (4 ways)

5(5 paise),  1(25paise)       (1 way)

10 and 25 can not be used alone

Total number of options = 9+4+1+4+1 = 19 ways 

With any three type coins

5(1 paise),  1(5 paise), 4(10  paise)
5(1 paise),  3(5 paise), 3(10  paise)
5(1 paise),  5(5 paise), 2(10  paise)
5(1 paise),  6(5 paise), 1(10  paise)
10(1 paise),  2(5 paise), 3(10  paise)
10(1 paise),  4(5 paise), 2(10  paise)
10(1 paise),  6(5 paise), 1(10  paise)
15(1 paise),  1(5 paise), 3(10  paise)
15(1 paise),  3(5 paise), 2(10  paise)
15(1 paise),  5(5 paise), 1(10  paise)
20(1 paise),  2(5 paise), 2(10  paise)
20(1 paise),  4(5 paise), 1(10  paise)
25(1 paise),  1(5 paise), 2(10  paise)
25(1 paise),  3(5 paise), 1(10  paise)
30(1 paise),  2(5 paise), 1(10  paise)
35(1 paise),  1(5 paise), 1(10  paise)   (total 16 ways)

5(1 paise),    4(5 paise), 1(25 paise)
10(1 paise),  3(5 paise), 1(25 paise)
15(1 paise),  2(5 paise), 1(25 paise)
20(1 paise),  1(5 paise), 1(25 paise)    (4 ways)

5(1 paise) 2(10 paise) 1(25 paise)
15(1 paise) 1(10 paise) 1(25 paise)     (2 ways)

1(5 paise),  2(10 paise), 1(25 paise)
3(5 paise),  1(10 paise), 1(25 paise)    (2 ways)

Total number of options = 16+4+2+2 = 24 ways

With all type coins
5(1 paise),  2(5 paise), 1(10 paise), 1(25 paise)
10(1 paise),  1(5 paise), 1(10 paise), 1(25 paise)

Total = 2 ways 

Required number of combinations = 4+19+24+2 = 49

Answer : Option A

Explanation:

No of ways of dividing n identical things among r different group

(for non empty set i.e, each one get at least one) is given by (n-1) C (r-1)

so on applying this to given situation we get 9C5