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**Q.1. Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.****[Hint: 2 women occupy the chair, from 1 to 4 in ^{4}P_{2} ways and 3 men occupy the remaining chairs in ^{6}P_{3} ways.]**

We have 2 women and 3 men

First women choose the chairs amongst the chairs 1 to 4

i.e. total number of chairs = 4

So, the number of arrangements =

Now 3 men choose from the remaining 6 chairs

So, the number of arrangements =

âˆ´ Total number of arrangements =

Hence, the total number of possible arrangements are 1440.

The alphabetical order of RACHIT is A, C, H, I, R and T

Number of words beginning with A = 5!

Number of words beginning with C = 5!

Number of words beginning with H = 5!

Number of words beginning with I = 5!

and Number of word beginning with R i.e. RACHIT = 1

âˆ´ The rank of the word â€˜RACHITâ€™ in the dictionary

= 5! + 5! + 5! + 5! + 1 = 4 x 5! + 1

= 4 x 5 . 4 . 3 . 2 . 1 + 1 = 4 x 120 + 1 = 480 + 1 = 481

Total number of questions = 12

Number of questions in each group = 6

7 questions are to be attempted but not more than 5 questions from either group

âˆ´ Total number of ways =

= 2[

2[6 x 15 + 15 x 20] = 2[90 + 300] = 2 x 390 = 780

Hence, the total number of ways = 780

Total number of points = 18

Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.

âˆ´ Total number of straight line formed by joining 2 points out of 18 points =

Number of straight lines formed by joining 2 points out of 5 points =

But 5 points are collinear and we get only one line when they are joined pairwise.

So, the required number of straight lines are

Hence, the total number of straight lines = 144.

Total number of persons = 8

Number of persons to be selected = 6

Condition is that if A is chosen, B must be chosen

Case I: When A is chosen, B must be chosen

Number of ways =

Case II: When A is not chosen, then B may be chosen

âˆ´ Number of ways =

So, the total number of ways =

[âˆ´ There are two cases]

=

Hence, the required number of ways = 22.

Total number of Persons = 12

Number of persons to be selected = 5

Out of 5, there is a chairperson

âˆ´ Number of ways of selecting a chairperson =

Number of ways of selecting other 4 numbers out of remaining 11 persons =

âˆ´ Total number of ways =

Hence, the required number of ways = 3960.

We have 26 English alphabet and 10 digits (0 to 9)

Since, it is given that each plate contains 2 different letters followed by 3 different digits.

âˆ´ Number of arrangement of 26 letters taken 2 at a time

Three digit number can be formed out of 10 digit =

âˆ´ Total number of license plates = 650 x 720 = 468000

Hence, the required number of plates = 468000.

Q.8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot.

Given that bag contains 5 black and 6 red balls.

Number of ways of selecting 2 black balls out of 5 black balls =

And number of ways of selecting 3 red balls out of 6 red balls =

âˆ´ Total number of ways of selecting 2 black and 3 red balls =

Hence, the required ways of selecting the balls = 200.

Total number of things = n 3 things must be together

âˆ´ The number of remaining things = n - 3

Number of things to be selected = r

Out of r, 3 are always together

âˆ´ Number of ways of selection =

Now permutation of 3 things which are always together = 3!

Number of permutations of (r â€“ 2) things = (r â€“ 2)!

âˆ´ Total number of arrangements =

Hence the required arrangements =

Total number of words in â€˜TRIANGLEâ€™ = 8

Out of 5 are consonants and 3 are vowels

If vowels are not together, taken we have the following arrangement

V | C | V | C | V | C | V | C | V | C | V

Consonant can be arranged in 5! = 120 ways

Vowel occupy 6 places

âˆ´ 3 vowels can be arranged in 6 places =

So, the total arrangement = 120 x 120 = 14400 ways

Here, the required arrangement = 14400 ways.

Any number divisible by 5, its unit place must have 0 or 5

We have to find 4-digit number greater than 6000 and less than 7000.

So, the unit place can be filled with 2 ways (0 or 5) since, repetition is not allowed

âˆ´ Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.

But the required number is greater than 6000 and less than 7000. So, thousand place can be filled with 1 digits i.e.6

So, the total number of integers = 1 x 8 x 7 x 2 = 112

Hence, the required number of integers = 112

Given that P

âˆ´ selection is to be done only for 10 - 3 = 7 persons

âˆ´ Number of selection =

5 people can be arranged as 5!

So, the number of arrangement = 35 x 5! = 35 x 120 = 4200

Hence, the required arrangement = 4200.

[Hint: Required number = 2

Total number of lamps = 10

The total number of ways in which hall can be illuminated is equal to the number of selection of one or more items out of n different items.

i.e.

From Binomial expansion, we have

So total number of ways =

= 2

Hence, the required number of possible ways = 1023.

We have 2 white, 3 black and 4 red balls in a box. 3 balls are to be drawn out of 9 balls atleast one black ball is to be included

So, the possible selection is

(1 black and 2 other balls) or (2 black and 1 other ball) or (3 black and no other ball)

So, the number of possible selection is

=

= 3 x 15 + 3 x 6 + 1 x 1 = 45 + 18 + 1 = 64

Hence, the required selection = 64.

Given that

Dividing eq. (i) by. (ii) we get

â‡’

â‡’ 3n â€“ 3r + 3 = 7r â‡’ 3n - 10r = - 3 ...(iv)

Now dividing eq. (ii) by eq. (iii), we get

â‡’ 2n â€“ 2r = 3r + 3 â‡’ 2n - 5r = 3 ...(v)

Solving eq. (iv) and (v) we have

â‡’ 2 x 9 - 5r = 3 â‡’ 18 - 5r = 3

â‡’

So

Hence, the value of

Given that all the 5 digit numbers are greater than 7000.

So, the ways of forming 5-digit numbers = 5 x 4 x 3 x 2 x 1 = 120

Now all the four digit number greater than 7000 can be formed as follows.

Thousand place can be filled with 3 ways

Hundred place can be filled with 4 ways

Tenths place can be filled with 3 ways

Units place can be filled with 2 ways

So, the total number of 4-digits numbers = 3 x 4 x 3 x 2 = 72

âˆ´ Total number of integers = 120 + 72 = 192

Hence, the required number of integers = 192

Given that out of 20 lines, no two lines are parallel and no three lines are concurrent.

Therefore, number of point of intersection

=

Hence, the required number of points = 190.

If first two digits is 41, then the remaining 4 digits can be arranged in

Similarly first two digits can be 42 or 46 or 62 or 64.

âˆ´ Total number of telephone numbers have all digits distinct = 5 x 1680 = 8400

Hence, the required telephone numbers = 8400

Given that question number 1 and 2 are compulsory.

âˆ´ The remaining questions are 5 - 2 = 3

Total number of questions to be attempted = 4 questions 1 and 2 are compulsory

So only 2 questions are to be done out of 3 questions Therefore number of ways =

[âˆ´

Hence, the required number of ways = 3.

Let n be the number of sides in a polygon.

Since, Polygon of n sides has (

= n

= (n â€“ 11) (n + 8) = 0 âˆ´ n = 11 and n = â€“ 8 [âˆµ n â‰ - 8]

So n = 11

Hence, the required number of sides = 11.

Given that 18 mice were placed equally in two experimental groups and one control group i.e. 3 groups

âˆ´ The required number of arrangements

Hence, the required arrangements =

Total number of marbles = 6 white + 5 red = 11 marbles

âˆ´ Required number of ways =

then the required number of ways =

then, the required number of ways =

Hence the required number of ways are

(i)

Given that the total number of players = 16 We have to select 11 players out of 16 players.

then then number of ways of selection = 16 â€“

then the number of ways of selection = 16 â€“

Hence, the required number of ways of selection

(i)

Total number of students in each class = 20

We have to select atleast 5 students from each class.

We have the following cases.

So, number of ways of selection of a team of 11 players =

= 2[

Hence, the required ways of selection = 2[

We have 4 girls and 7 boys and a team of 5 members is to be selected.

i.e.

Number of ways =

= 4 x 35 + 6 x 35 + 4 x 21 + 7 = 140 + 210 + 84 + 7 = 441

ways

Number of ways =

Hence the required number of ways are

(i) 21 ways (ii) 441 ways (iii) 91 ways

Given that

âˆ´ n â€“ 8 = 12 â‡’ n = 12 + 8 = 20

Hence, the correct option is (a)

We know that a coin has Head and Tail (H, T)

âˆ´ When a coin is tossed 6 times, then the

Possible outcome = 26 = 64

Hence, the correct option is (b).

Four-digit numbers are to be formed from the digits 2, 3, 4, 7 without repetition

So, the required 4-digit numbers =

Hence, the correct option is (c).

If we fix 3 at unit place, then the total possible numbers = 3!

If we fix 4, 5 and 6 at unit place, this is each case, total possible numbers are 3!

Required sum of unit digits of all such numbers is

= 3 x 3! + 4 x 3! + 5 x 3! + 6 x 3! = (3 + 4 + 5 + 6) x 3!

= 18 x 3! = 18 x 3 x 2 x 1 = 108

Hence, the correct option is (b).

Given that total numbers of vowels = 4

and total numbers of consonants = 5

Total number of words formed by 2 vowels and 3 consonants

= 6 x 10 = 60

Now permutation of 2 vowels and 3 consonants = 5!

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

So, the total number of words = 60 x 120 = 7200.

Hence, the correct option is (c).

We know that a number is divisible by 3 when the sum of its digits is divisible by 3.

If we take the digits 0, 1, 2, 4, 5, then the sum of the digits

= 0 + 1 + 2 + 4 + 5 = 12 which is divisible by 3

So, the 5 digit numbers using the digits 0, 1, 2, 4, and 5

4 x 4 x 3 x 2 x 1 = 96

and if we take the digits 1, 2, 3, 4, 5, then their sum

= 1 + 2 + 3 + 4 + 5 = 15 divisible by 3

So, five digit numbers can be formed using the digits 1, 2, 3, 4, 5 is 5! ways = 5 x 4 x 3 x 2 x 1 = 120 ways

Total number of ways = 96 + 120 = 216

Hence, the correct option is (a).

Q.32. Every body in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is

Let the total number of persons in a room be n since, two persons make 1 hand shake

âˆ´ The number of hand shakes =

So

â‡’

â‡’

â‡’ n

â‡’ n(n â€“ 12) + 11 (n â€“ 12) = 0 â‡’ (n - 12) (n + 11) = 0

â‡’ n â€“ 12 = 0, n + 11 = 0â‡’ n = 12, n = â€“ 11

âˆ´ n = 12 (âˆ´ n â‰ - 11)

Hence, the correct option is (b).

Total number of triangles formed from 12 points taking 3 at a time = 12C3

But given that out of 12 points, 7 are collinear

So, these seven points will form no triangle.

âˆ´ The required number of triangles =

Hence, the correct option is (d).

We know that to form a parallelogram, we require a pair of lines from a set of 4 lines and another pair of lines from another set of 3 lines

âˆ´ Required numbers of parallelograms =

Hence, the correct option is (b).

Total number of players = 22

2 players are always included and 4 are always excluding or never included = 22 â€“ 2 â€“ 4 = 16

âˆ´ Required number of selection =

Hence, the correct option is (c).

Total number of 5-digit telephone number if all the digits are repeated = (10)

If digits are not repeated, then 5-digit telephones, can be formed in

âˆ´ Required number of ways =(10)

= 100000 â€“ 30240 = 69760

Hence, the correct option is (d).

Number of men = 4

Number of women = 6

We are given that the committee includes 2 men and exactly twice as many women as men.

Thus, the possible selection can be 2 men and 4 women and 3 men and 6 women.

So, the number of committee =

= 6 x 15 + 4 x 1 = 90 + 4 = 94

Hence, the correct option is (a).

We have to form 9 digit numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

and we know that 0 can not be put on extremely left place.

So, first place from the left can be filled in 9 ways.

Now repetition is not allowed. So, the remaining 8 places can be filled in 9!

âˆ´ The required number of ways = 9 x 9!

Hence, the correct option is (c).

Total number of letters in the â€˜ARTICLEâ€™ is 7 out which A, E, I are vowels and R, T, C, L are consonants

Given that vowels occupy even place

âˆ´ possible arrangement can be shown as below

C, V, C, V, C, V, C i.e. on 2nd , 4th and 6th places

Therefore, number of arrangement =

Now consonants can be placed at 1, 3, 5 and 7th place

âˆ´ Number of arrangement =

So, the total number of arrangements = 6 x 24 = 144

Hence, the correct option is (b).

Possible number of choosing 5 different green dyes = 2

Possible number of choosing 4 blue dyes = 2

and possible number of choosing 3 red dyes = 2

If atleast one blue and one green dyes are selected then the

total number of selection = (2

Hence, the correct option is (b).

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