Answer: Option B. -> 48
There are 2 primary teachers.
They can stand in a row in
P (2, 2) = 2! = 2 × 1 ways = 2 ways
∴ Two middle teachers.
They can stand in a row in
P (2, 2) = 2! = 2 × 1 ways = 2 ways
There are two secondary teachers.
They can stand in a row in
P (2, 2) = 2!= 2 × 1 ways = 2 ways
These three sets can be arranged themselves in
3! ways = 3 × 2 × 1 = 6 ways
Hence,, the required number of ways
= 2 × 2 × 2 × 6
= 48 ways

Answer: Option D. -> 362880
The given word contains 9 letters, all different.
∴ Required number of ways
$$\eqalign{
& = {}^9{P_9} \cr
& = 9! \cr
& = \left( {9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) \cr
& = 362880 \cr} $$

Answer: Option D. -> 63
Number of ways of attempting 1st, 2nd, 3rd question are each.
Total number of ways
43 = 4 × 4 × 4 = 64
Number of ways, getting correct answers = 13 = 1
∴ Number of ways of not getting all answer correct = 64 - 1 = 63

Answer: Option C. -> 4 × 34
The remainder on the first card can be 0, 1, 2 or 3 i.e. 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).
For each value on the card the remainder can have 3 possible values.
The total number of possible sequences is: 4 × 34

Answer: Option B. -> 12C4 × 4C3 × 7!
4 consonants out of 12 can be selected in,12C4 ways.
3 vowels can be selected in 4C3 ways.
Therefore, total number of groups each containing 4 consonants and 3 vowels, = 12C4 × 4C3
Each group contains 7 letters, which can be arranging in 7! ways.
Therefore required number of words, = 12C4 × 4C3 × 7!

Answer: Option B. -> 360
Two horses A and B, in a race of 6 horses . . . A has to finish before B.
If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4!
If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4!
If A finishes 3rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4!
If A finishes 4th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4!
If A finishes 5th . . . . B has to be 6th and the top 4 positions could be filled in 4! ways.
A cannot finish 6th, since he has to be ahead of B.
Therefore total number of ways:
= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!
= 120 + 96 + 72 + 48 + 24
= 360

Answer: Option D. -> 91
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 in 5+2C1 = 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 9C3 = 84 ways.
Thus, total number of ways is 7 + 84 = 91 ways.

Answer: Option B. -> 3
Let the number of Rose plants be a.
Let number of marigold plants be b.
Let the number of Sunflower plants be c.
According to question,
20a + 5b + 1c = 1000 - - - - - - (1)
a + b + c = 100 - - - - - - - - - - (2)
Solving the above two equations by eliminating c,
19a + 4b = 900
b = $$\frac{{900 - 19a}}{4}$$   = $$225 - \frac{{19a}}{4}$$   - - - - - - - (3)
b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .
0 < b < 99 - - - - - - - (4)
Substituting (3) in (4),
0 < 225 - $$\frac{{19a}}{4}$$ < 99
⇒ 225 < -$$\frac{{19a}}{4}$$ < (99 - 225)
⇒ 4 × 225 > 19a > 126 × 4
⇒ $$\frac{{900}}{{19}}$$ > a > 504
a is the integer between 47 and 27 - - - - - - - - (5)
From (3), it is clear, a should be multiple of 4.
Hence, possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.

Answer: Option D. -> 756
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
∴ the Required number of ways
$$\eqalign{
& = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + \left( {{}^7{C_4} \times {}^6{C_1}} \right) + \left( {{}^7{C_5}} \right) \cr
& = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {{}^7{C_3} \times {}^6{C_1}} \right) + \left( {{}^7{C_2}} \right) \cr
& = 525 + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right) + \left( {\frac{{7 \times 6}}{{2 \times 1}}} \right) \cr
& = \left( {525 + 210 + 21} \right) \cr
& = 756 \cr} $$

Answer: Option C. -> 25200
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
$$\eqalign{
& = \left( {{}^7{C_3} \times {}^4{C_2}} \right) \cr
& = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{4 \times 3}}{{2 \times 1}}} \right) \cr
& = 210 \cr} $$
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!
= 5 x 4 x 3 x 2 x 1
= 120
∴ Required number of ways = (210 x 120) = 25200