Explore topic-wise InterviewSolutions in .

Given as 

Multiple choice question, only one answer is correct of the given options.

For the first three questions only one answer is correct out of four. Therefore it can be answered in 4 ways.

The total number of ways to answer the first 3 questions = ⁴C₁ × ⁴C₁ × ⁴C₁ = 4 × 4 × 4 = 64

Each of the next 3 questions can be answered in 2 ways.

The total number of ways to answer the next 3 questions = ²C₁ × ²C₁ × ²C₁ = 2 × 2 × 2 = 8

Thus, total possible outcomes possible are 64 × 8= 512

(i) Given as there are 5 books of mathematics and 6 books of physics.

Now, in order to buy one mathematics book, number of ways is ⁵C₁ similarly to buy one physics book number of ways is ⁶C₁

Thus, the number of ways a student buy a Mathematics book and a Physics book is ⁵C₁ × ⁶C₁ = 5 × 6 = 30

(ii) Given as there is a total of 11 books.

Therefore in order to buy either a Mathematics book or a Physics book it means that only one book out of eleven is bought.

Thus, the number of ways in which a student can either buy either a Mathematics book or a Physics book is ¹¹C₁ = 11

Given as

The digits 1, 2, 3, 4, 3, 2, 1

Total number of digits are 7.

Here's 4 odd digits 1,1,3,3 and 4 odd places (1,3,5,7)

Therefore, the odd digits can be arranged in odd places in n!/ (p! × q! × r!) = 4!/(2! 2!) ways.

And the remaining even digits 2,2,4 can be arranged in 3 even places in n!/ (p! × q! × r!) = 3!/2! Ways.

Thus, the total number of digits = 4!/(2! 2!) × 3!/2!

= [4 × 3 × 2 × 1 × 3 × 2 × 1] / (2! 2! 2!)

= 3 × 2 × 1 × 3 × 1

= 18

Thus, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.

A person has 5 chocolates of different sizes and they are to be distributes among 5 children of different ages. If the biggest chocolate is to be given to the youngest child, then remaining 4 chocolates can be distributed among the remaining 4 children in ⁴P₄ ways.

∴ Total permutations = ¹P₁ x ⁴P₄

= 1 × 4!

= 1 × 24

= 24

Given as

The number of roots from Goa to Bombay is air and sea.

Therefore, the number of ways to go from Goa to Bombay is ²C₁

Given as the number of roots from Bombay to Delhi are: air, rail, and road.

Therefore, the number of ways to go from Bombay to Delhi is ³C₁

Thus, the number of ways to go from Goa to Delhi via Bombay is ²C₁ × ³C₁ = 2 × 3 = 6 ways.

Given: 5 routes from A to B and 3 routes from B to C. 

To find: number of different routes from A to C via B. 

Let E₁ be the event : 5 routes from A to B 

Let E₂ be the event : 3 routes from B to C 

Since going from A to C via B is only possible if both the events E₁ and E₂ occur simultaneously. 

So there are 5 × 3 =15 different routes from A to C via B.

To Prove : (n+2) x (n!) + (n+1)! = (n!) x (2n + 3)

Formula : n! = n x (n-1)!

L.H.S. = (n + 2) × (n!) + (n + 1)! 

= (n + 2) × (n!) + (n + 1) × (n!) 

= (n!) × [(n + 2) + (n + 1)] 

= (n!) × (2n + 3) 

= R.H.S. 

∴L.H.S. = R.H.S. 

Conclusion : (n + 2) × (n!) + (n + 1)! = (n!) × (2n + 3)

To find: number of ways in which 5 ladies draw water from 5 taps. 

Condition: no tap remains unused 

The condition given is that no well should remain unused. 

So possible number of ways are: 5 × 4 × 3 × 2 × 1=120.

A bulb can be good or defective, so there are 2 different possibilities of a bulb. 

So number of all possible outcomes (of all bulbs) = 2 x 2 x 2 = 8

Given: 20 students. 

The number of ways of giving first and second prizes in mathematics to a class of 20 students = 20 x 19. 

(First prize can be given to any one of the 20 students but the second prize cannot be given to the student that received the first prize so the number of candidates for the second prize is 19.) 

The number of ways of giving first and second prizes in chemistry 

To a class of 20 students = 20 × 19. 

The number of ways of giving first prize in physics to a class of 20 students = 20 

The number of ways of giving first prize in English to a class of 20 students = 20 

So total number of ways = 20 × 19 × 20 × 19 × 20 × 20 = 57760000

Given: a set of five true – false questions. 

To find: the maximum number of students in the class.

Condition: no student has written the all correct answer and no two students have given the same sequence of answers. 

The total number of answering a set of 5 true or false question = 2⁵ = 32 

Since, no two students have given the same sequence of answers and no student has written the all correct answer. 

Therefore total possibilities reduces by 1(of no student has written the all correct answer) 

⇒2⁵ -1 = 32 - 1 = 31

Given Equation : 

(n + 1)! = 12 × (n - 1)! 

To Find : Value of n 

Formula : n! = n x (n-1)!

By given equation,

(n + 1)! = 12 × (n - 1)! 

By using above formula we can write, 

∴(n + 1) × (n) × (n - 1)! = 12 × (n - 1)! 

Cancelling the term (n - 1)! from both the sides,

∴(n + 1) × (n) = 12 …….. eq(1) 

∴(n + 1) × (n) = (4) × (3) 

Comparing both the sides, we get, 

∴n = 3

Conclusion : Value of n is 3

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 12 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

(i) To distribute 3 prizes among 4 girls where no girl gets more than one prize the possible number of permutation possible are:⁴P₃ = 24

(ii) To distribute 3 prizes among 4 girls where a girl may get any number of prizes the number of possibilities are:4 × 4 × 4 = 64. (Since a prize can be given to any of the 4 girls.) 

(iii) To distribute 3 prizes among 4 girls where no girl gets all the prizes the number of possibilities are: (4 × 4 × 4)-(4) = 64 - 4 = 60 

(The situation where a single girl gets all the prizes has to reduced from the situation where a girl may get any number of prizes.)

We have to find the possible number of ways in which we can put twenty balls in five boxes so that the first box contains only one ball when repetition of distribution of balls is allowed. 

We will use the concept of multiplication because there are twenty sub jobs dependent on each other and are performed one after the other. 

The thing that is distributed is considered to have choices not the things to which we have to give them, it means that in this problem the balls have choices more precisely four choices are there for each ball and boxes won’t choose any because letters have the right to choose. But one box has the right to choose any one of the twenty balls, so the first box has twenty choices. 

The number of ways in which we can put nineteen balls in four boxes where repetition of distribution is allowed 

4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4¹⁹ 

Hence,

The total number of ways in which we can put twenty balls in five boxes such that first box contains only one ball = 20 × 4¹⁹.

ILLUSTRATION – 

To illustrate suppose there are five balls a, b, c, d, e to be put in three boxes A, B, C such that first box contains only one ball. 

Let the first box be A and suppose we have combination of a, A , so the rest combinations would be 2× 2 × 2 × 2 = 2⁴, but we have more combinations like b, A ; c, A ; d, A ; e, A these combinations are five , so we add them at the end, and the final answer we get is = 5 × 2⁴.

We have to find the possible number of ways in which we can put five balls in three boxes when repetition of distribution of balls is allowed. 

We will use the concept of multiplication because there are five sub jobs dependent on each other and are performed one after the other. 

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the balls have choices more precisely three choices are there for each ball and boxes won’t choose any because balls have the right to choose. 

The number of ways in which we can put five balls among three boxes where repetition of distribution is allowed 3 × 3 × 3 × 3 × 3 

= 3⁵ 

= 243.

Given as

The digits 2, 3, 0, 3, 4, 2, 3

The total number of digits = 7

As we know that, zero cannot be the first digit of the 7 digit numbers.

The number of 6 digit number = n!/ (p! × q! × r!) = 6! / (2! 3!) Ways. [2 is repeated twice and 3 is repeated 3 times]

Total number of arrangements = 6! / (2! 3!)

= [6 × 5 × 4 × 3 × 2 × 1] / (2 × 3 × 2)

= 5 × 4 × 3 × 1 

= 60

Then, number of 7 digit number = n!/ (p! × q! × r!) = 7! / (2! 3!) Ways

Total number of arrangements = 7! / (2! 3!)

= [7 × 6 × 5 × 4 × 3 × 2 × 1] / (2 × 3 × 2)

= 7 × 5 × 4 × 3 × 1

= 420

Therefore, total numbers which is greater than 1 million = 420 – 60 = 360

Thus, total number of arrangements of 7 digits (2, 3, 0, 3, 4, 2, 3) forming a 7 digit number is 360.

Correct option is (d) 4

Given as

P (9, r) = 3024

On using the formula,

P (n, r) = n!/(n – r)!

P (9, r) = 9!/(9 – r)!

Therefore, from the question,

P (9, r) = 3024

On substituting the obtained values in above expression we get,

9!/(9 – r)! = 3024

1/(9 – r)! = 3024/9!

= 3024/(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 3024/(3024 × 5 × 4 × 3 × 2 × 1)

= 1/5!

(9 – r)! = 5!

9 – r = 5

-r = 5 – 9

-r = -4

Hence, the value of r is 4.

Given as 

P (11, r) = P (12, r – 1)

On using the formula,

P (n, r) = n!/(n – r)!

P (11, r) = 11!/(11 – r)!

P (12, r-1) = 12!/(12 – (r-1))!

= 12!/(12 – r + 1)!

= 12!/(13 – r)!

Therefore, from the question,

P (11, r) = P (12, r – 1)

On substituting the obtained values in above expression we get,

11!/(11 – r)! = 12!/(13 – r)!

Now, upon evaluating,

(13 – r)! / (11 – r)! = 12!/11!

[(13 – r) (13 – r – 1) (13 – r – 2)!] / (11 – r)! = (12×11!)/11!

[(13 – r) (12 – r) (11 -r)!] / (11 – r)! = 12

(13 – r) (12 – r) = 12

156 – 12r – 13r + r² = 12

156 – 12 – 25r + r² = 0

r² – 25r + 144 = 0

r² – 16r – 9r + 144 = 0

r(r – 16) – 9(r – 16) = 0

(r – 9) (r – 16) = 0

r = 9 or 16

For, P (n, r): r ≤ n

∴ r = 9 [for, P (11, r)]

As we know that in odd numbers, the last digit consists of (1, 3, 5, 7, 9).

Let us assume that we have three boxes.

First box can be filled with any one of the nine digits (zero not allowed at first position)

Therefore the possibilities are ⁹C₁

The second box can be filled with any one of the ten digits

Therefore the available possibilities are ¹⁰C₁

Third box can be filled with any one of the five digits (1,3,5,7,9)

Therefore the available possibilities are ⁵C₁

Thus, the total number of possible outcomes are ⁹C₁ × ¹⁰C₁ × ⁵C₁ = 9 × 10 × 5 = 450

Given as

P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7

P(2n – 1, n) / P(2n + 1, n – 1) = 22/7

On using the formula,

P (n, r) = n!/(n – r)!

P (2n – 1, n) = (2n – 1)! / (2n – 1 – n)!

= (2n – 1)! / (n – 1)!

P (2n + 1, n – 1) = (2n + 1)! / (2n + 1 – n + 1)!

= (2n + 1)! / (n + 2)!

Therefore, from the question,

P(2n – 1, n) / P(2n + 1, n – 1) = 22/7

On substituting the obtained values in above expression we get,

[(2n – 1)! / (n – 1)!] / [(2n + 1)! / (n + 2)!] = 22/7

[(2n – 1)! / (n – 1)!] × [(n + 2)! / (2n + 1)!] = 22/7

[(2n – 1)! / (n – 1)!] × [(n + 2) (n + 2 – 1) (n + 2 – 2) (n + 2 – 3)!] / [(2n + 1) (2n + 1 – 1) (2n + 1 – 2)] = 22/7

[(2n – 1)! / (n – 1)!] × [(n + 2) (n + 1) n(n – 1)!] / [(2n + 1) 2n (2n – 1)!] = 22/7

[(n + 2) (n + 1)] / (2n + 1)2 = 22/7

7(n + 2) (n + 1) = 22×2 (2n + 1)

7(n² + n + 2n + 2) = 88n + 44

7(n² + 3n + 2) = 88n + 44

7n² + 21n + 14 = 88n + 44

7n² + 21n – 88n + 14 – 44 = 0

7n² – 67n – 30 = 0

7n² – 70n + 3n – 30 = 0

7n(n – 10) + 3(n – 10) = 0

(n – 10) (7n + 3) = 0

n = 10, -3/7
As we know that, n ≠ -3/7

∴ The value of n is 10.

Let us assume that we have three boxes.

First box can be filled with any one of the nine digits (zero not allowed at first position)

Therefore, possibilities are ⁹C₁

The second box can be filled with any one of the ten digits

Therefore the available possibilities are ¹⁰C₁

Third box can be filled with any one of the ten digits

Therefore the available possibilities are ¹⁰C₁

Thus, the total number of possible outcomes are ⁹C₁ × ¹⁰C₁ × ¹⁰C₁ = 9 × 10 × 10 = 900.

Given as

P(n, 5) : P(n, 3) = 2 : 1

P(n, 5) / P(n, 3) = 2/1

On using the formula,

P (n, r) = n!/(n – r)!

P (n, 5) = n!/ (n – 5)!

P (n, 3) = n!/ (n – 3)!

Therefore, from the question,

P (n, 5) / P(n, 3) = 2/1

On substituting the obtained values in above expression we get,

[n!/ (n – 5)!] / [n!/ (n – 3)!] = 2/1

[n!/ (n – 5)!] × [(n – 3)! / n!] = 2/1

(n – 3)! / (n – 5)! = 2/1

[(n – 3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 2/1

[(n – 3) (n – 4) (n – 5)!] / (n – 5)! = 2/1

(n – 3)(n – 4) = 2

n² – 3n – 4n + 12 = 2

n² – 7n + 12 – 2 = 0

n² – 7n + 10 = 0

n² – 5n – 2n + 10 = 0

n (n – 5) – 2(n – 5) = 0

(n – 5) (n – 2) = 0

n = 5 or 2

For, P (n, r): n ≥ r

∴ n = 5 [for, P (n, 5)]

Given : 

The three-digit number is required without digit repetition Assume we have three boxes; the first box can be filled with any one of the nine digits (0 not allowed at first place). 

Therefore,

Possibilities are ⁹C₁, the second box can be filled with any one of the nine digits available possibilities are ⁹C₁, the third box can be filled with any one of the eight digits available possibilities are ⁸C₁. 

Hence,

Number of total outcomes possible are ⁹C₁ × ⁹C₁ × ⁸C₁ 

= 9 × 9 × 8 

= 648

Given : A total of three positions are to be given. Number of ways to select principal is ³⁶C₁ 

Number of ways to select vice-principal is ³⁵C₁ 

(35 since one position is already given) 

Number of ways to select teacher in charge is ³⁴C₁ 

(34 since two positions are already given) 

Hence,

Total number of ways is ₃₆C¹ × ₃₅C¹ × ₃₄C¹ 

= 36 × 35 × 34 

= 42840

Given as

P(15, r – 1) : P(16, r – 2) = 3 : 4

P(15, r – 1) / P(16, r – 2) = 3/4

On using the formula,

P (n, r) = n!/(n – r)!

P (15, r – 1) = 15! / (15 – r + 1)!

= 15! / (16 – r)!

P (16, r – 2) = 16!/(16 – r + 2)!

= 16!/(18 – r)!

Therefore, from the question,

P(15, r – 1) / P(16, r – 2) = 3/4

On substituting the obtained values in above expression we get,

[15! / (16 – r)!] / [16!/(18 – r)!] = 3/4

[15! / (16 – r)!] × [(18 – r)! / 16!] = 3/4

[15! / (16 – r)!] × [(18 – r) (18 – r – 1) (18 – r – 2)!]/(16 × 15!) = 3/4

1/(16 – r)! × [(18 – r) (17 – r) (16 – r)!]/16 = 3/4

(18 – r) (17 – r) = 3/4 × 16

(18 – r) (17 – r) = 12

306 – 18r – 17r + r² = 12

306 – 12 – 35r + r² = 0

r² – 35r + 294 = 0

r² – 21r – 14r + 294 = 0

r(r – 21) – 14(r – 21) = 0

(r – 14) (r – 21) = 0

r = 14 or 21

For, P(n, r): r ≤ n

Hence r = 14 [for, P(15, r – 1)]

This is the example of Cartesian product of two sets. 

The pairs in which the first entry is an element of A and the second is an element of B are : 

(2,0),(2,1),(3,0),(3,1),(5,0),(5,1) 

⇒ 3 × 2 = 6

Given: Two sets: {1, 2, 3} & {2, 3, 4} 

To find: number of A.P. with 10n terms whose first term is in the set {1, 2, 3} and whose common difference is in the set {2, 3, 4} 

Number of arithmetic progressions with 10 terms whose first term are in the set {1, 2, 3} and whose common difference is in the set {2, 3, 4} are: 3 × 3 = 9 

(3 because there are three elements in the set {1, 2, 3} and another 3 because there are three elements in the set {2, 3, 4})

To Find: Value of n 

Given: (n+1)! = 12× [(n-1)!] 

Formula Used: n! = (n) × (n-1) × (n-2) × (n-3) ………. 3 × 2 × 1

Now, (n+1)! = 12× [(n-1)!] 

⇒ (n+1) × (n) × [(n-1)!] = 12 × [(n-1)!] 

⇒ (n+1) × (n) = 12 

⇒ n²+n = 12

 ⇒ n²+n-12 = 0 

⇒ (n-3) (n+4) = 0 

⇒ n = 3 or, n = -4 

But, n=-4 is not possible because in case of factorial (!) n cannot be negative. Hence, n=3 is the correct answer.

To Find: Value of n

Given: \(\frac{1}{4!} +\frac{1}{5!} = \frac{x}{6!}\) 

Formula Used: n! = (n) × (n-1) × (n-2) × (n-3) ………. 3 × 2 × 1

Now, \(\frac{1}{4!} +\frac{1}{5!} = \frac{x}{6!}\) 

⇒ \(\frac{1}{24} +\frac{1}{120} = \frac{x}{720}\) (4! = 24, 5! = 120)

⇒ \(\frac{5+1}{120}=\frac{x}{720}\)

⇒ \(\frac{6}{120}=\frac{x}{720}\)

⇒ x = 36

Given,

The word LATE. 

Arranging the permutations of the letters of the word LATE in a dictionary : 

To find : Rank of word LATE in dictionary.

First comes,

words starting with letter A = 3! (3 letters, no repetation) 

Words starting with letter E = 3! (3 letters, no repetation) 

Words starting with L words starting with LA : 

LAET = 1 

LATE = 1 

Rank of the word LATE = 6 + 6 + 1 + 1 

= 14 

Hence,

The rank of the word LATE in arranging the letters of LATE in a dictionary among its permutations is 14.

There are 7 letters in the word HEXAGON. 

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)! 

Therefore, a permutation of 7 different objects in 7 places is 

P(7,7) = \(\frac{7!}{(7-7)!}\)  =  \(\frac{7!}{0!}\) = \(\frac{5040}{1}\) = 5040. 

They can be permuted in P (7,7) = 5040 ways.

The vowels in the word are E, A, O.

Consider this as a single group.

Now considering vowels as a single group, there are total 5 groups (4 letters and 1 vowel group) can be permuted in P (5,5) 

Now vowel can be arranged in 3! Ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)! 

Therefore, the arrangement of 5 groups and vowel group is

P(5,5)×3! =\(\frac{5!}{(5-5)!}\) × 3! = \(\frac{5!}{0!}\) x 3! = \(\frac{120}{1}\) x 6 = 720. 

Hence total number of arrangements possible is 720.

Except 3 small cages In remaining 4 cages 3 lions can be caged In ⁴P₃ ways.

Now, In the remainIng (1 big + 3 small) 4 cages the remainIng 4 lions can be caged In ⁴P₄ ways.

∴ Total permutations of cagIng 7 lions in

7 cages = ⁴P₃ . ⁴P₄

= (4 × 3 × 2) × 4!

= 24 × 24

= 576

Correct option is (a) n

Correct option is (b) 4

Correct option is (b) ⁿC_n-r

Correct option is (c) 2

Given: 5 objective-type question, each question having 4 choices. 

To find: the number of ways of answering them. 

Each objective-type question has 4 choices.

So the total number of ways of answering 5 objective-type question, each question having 4 choices = 4 × 4 × 4 × 4 × 4 = 4⁵