Answer : Option D

Explanation:

Here in the 3-digit number, in which two consecutive digits are same like 223, 551, 500 and so on ..... but not 111, 222 ... as it is mentioned exactly 2 digits same not 3 digits.

Case I:

Let us assume that the $2^{\mathrm{nd}}$ and $3^{\mathrm{rd}}$ positions are same, then it can be done in 10 ways like given below:

$1^{\mathrm{st}}$ position can take values from 1 to 9 only ( zero cannot be occupy).Therefore total ways comes out to b 9 x 10 = 90 ways.

Case I:

Let us assume that the  $1^{\mathrm{st}}$ and  $2^{\mathrm{nd}}$ positions are same as '0' cannot be occupied, then it can be done in 9 ways like given below:

$3^{\mathrm{rd}}$ position can take values from 1 to 10 (zero can occupy).Therefore total ways comes out to b 9 x 10 = 90 ways.

Total = 90 + 90 = 180 numbers.

Now remove these 9 common numbers like 111, 222, 333 ...... 999) from both cases.

Therefore finally 180 - 18 = 162

Answer : Option C

Explanation:

Here 8 coins are given:

5 rupee, 1 rupee, 50 paise, 25 paise, 10 paise, 3 paise, 2 paise and 1 paise.

Pick 1 coin from the above 8 coins, then we get 8 different values 5 rupee or 1 rupee or 50 paise or 25 paise or 10 paise or 3 paise or 2 paise or 1 paise = 8C1= 8 ways

Pick 2 coins from the above 8 coins = 8C2= 28 ways

We can get

Rs.5 + Rs.1 = Rs.6

Rs.5 + 50 paise = Rs.5.50

Rs.5 + 25 paise = Rs.5.25

 .

 Rs.1 + 10 paise = Rs.1.10 and so on.

Pick 3 coins from the above 8 coins = 8C3 = 56 ways

We can get

Rs.5 + Rs.1 + 50 paise = Rs.6.50

Rs.5 + Rs.1 + 25 paise = Rs.6.25

Rs.1 + 50 paise + 10 paise = Rs.1.60...so on.

...

Pick 4 coins

....

Pick 5 coins ...... Pick all 8 coins = 8C8 = 1 way

Hence the answer is 

= 8C1 + 8C2 + 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8 

= 255 ways 

Answer : Option D

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

Now 0 is not available, so 5 can occupy the last place.

           2,  3,  4,  5,  6,  7

            _3_    _4_    _5_

            _3_    _7_    _5_

            _4_    _3_    _5_

            _4_    _6_    _5_

            _6_    _4_    _5_

            _6_    _7_    _5_

            _7_    _3_    _5_

            _7_    _6_    _5_

Hence the total 8 numbers can be formed.

Answer : Option A

Explanation:

There are different cases involved in these

9 digit numbers

Do not remove any digit, it will be increasing order

(123456789)

8 digit numbers

Remove any 1 digit from the given 9 digit

Here we can remove any 1 digit from these 9 digits =  9C1= 9 ways 

 7 digit numbers

Remove any 2 digit from the given 9 digit

Here we can remove any 2 digit from these 9 digits =  9C2= 36 ways 

In the same manner, we can remove any 3 digits, 4 digits, 5 digits and so on ......  we can remove till 7 digits.

= 9C0+9C1+9C2+9C3+9C4+9C5+9C6+9C7  

= 1 + 9 + 36 + 84 + 126 + 126 + 84 + 36  

=  502 Numbers

Answer : Option B

Explanation:

Suppose lady has A, B, C, D, E, F, G, H, I, J are 10 friends.

Let us say A and B are enemies and not willing to come together.

Case I: If A is interested then B is not interested, then

 choosing 6 guests form 9 person can be done

A, C, D, E, F, G, H, I, J = ${\text{9}}_{{\mathrm{c}}_6}$= 84 ways

Case II: If B is interested then A is not interested, then

 choosing 6 guests form 9 person can be done

A, C, D, E, F, G, H, I, J = ${\text{9}}_{{\mathrm{c}}_6}$= 84 ways.

But remember, there are 8 people like C, D, E, F, G, H, I, J ( who are repeated in both cases, So remove them) = $8_{{\mathrm{c}}_6}$= 28 ways.

So the final combination is ${\text{9}}_{{\mathrm{c}}_6}\text{ + }{\text{9}}_{{\mathrm{c}}_6}{\text{ - 8}}_{{\mathrm{c}}_6}$ = 140 ways

Answer : Option A

Explanation:

 Video Explanation is given

Answer : Option D

Explanation:

For the six digit numbers using 0, 1, 2, 3, 4 and 5

0 cannot occupy the first and last position as the 6-digit number cannot be formed.

Answer : Option D

Explanation:

From 3 red balls and 2 white balls.

Keep 1 white ball fix in the end.

then, there are four balls left (i.e) 3 red and 1 white balls left

resulting 4 combinations

W R R R

R W R R

R R W R

R R R W

4 cases x  $\frac{1}{10}$ =  $\frac{2}{5}$

Answer : Option C

Explanation:

In this question, he asked about atleast 2 digits same (i.e) we can have more than 2 digits also.

For any number formed, we have possibilities:

No digit same + 2 digits same + 3 digits same + 4 digits same = Total numbers formed.

For example:

1234 , 1123,  1113,  1111

we don't need 1234 kind of numbers, and the rest of all type of numbers are accepted.

Therefore, it is possible, if we subtract

Total possible numbers with repetition - Without repetition   

So atleast 2 digits same = With repetition - Without repetition

= 9000 - 4536

= 4464

Answer : Option D

Explanation:

In this question, it is asked about four digit number plate, in which first two digits are alphabets ( letters ) and the next two are digits.

Remember : It is also said that repetition of digits is not allowed ( means only digits not alphabets 

Available items are 

Therefore the total number of number plates formed is 60840