Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14

= C 4 14 ways =  14 × 13 × 12 × 11 4 × 3 × 2 × 1= 1001

Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

   ? C 1 5 + C 4 4 + C 1 5 = 5+1+5 =11

 P(E) =  n ( E ) n ( S ) = 11 1001 = 1 91  

 Required probability =  1 - 1 91 = 90 91

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.

So, there is one way of doing it.

Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) =(  C 3 7 × C 2 4 ) = 210.

Number of groups, each having 3 consonants and 2 vowels =210

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120. 

Required number of words = (210 x 120) = 25200.

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

(or)

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.

The following arrangements satisfy all 3 conditions.

Arrangement 1: 3 books in a row; 12 rows.

Arrangement 2: 4 books in a row; 9 rows.
Arrangement 3: 6 books in a row; 6 rows.
Arrangement 4: 9 books in a row; 4 rows.
Arrangement 5: 12 books in a row; 3 rows.

Therefore, the possible arrangements are 5.

Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are 

5 x 4 x 3 x 2 x 1 = 120.

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). 

Required number of ways =  6 C 1 * 4 C 3 + 6 C 2 * 4 C 2 + 6 C 3 * 4 C 1 + 6 C 4  

=  6 C 1 * 4 C 1 + 6 C 2 * 4 C 2 + 6 C 3 * 4 C 1 + 6 C 2 = 209.

There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.

But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore, the total number of ways in which 8 students can travel is:
8 C 3 + 8 C 4=56 + 70= 126

We first count the number of committee in which

(i). Mr. Y is a member
(ii). the ones in which he is not

Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

We can choose 1 more in5+ 2 C 1=7 ways.

Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in 9 C 3=84 ways.

Thus, total number of ways is 7+84= 91 ways.