Topic Question Set


Q 11 :

The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is _______ .           [2023]



(240)

Given digits, 1, 2, 2, 2, 3, 3, 5

Total digits = 7

Total number of seven digits =7!3!2!=420

Total number of seven digit even numbers =3×6!2!3!=180

   Total number of seven digit odd numbers = 420 - 180 = 240



Q 12 :

Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _______ .              [2023]



(2997)

Number starting with 2 and 3 = 2×64

Number starting with 40 = 63

Number starting with 420, 422, 423, 424, 427 = 5×62

Similarly, number starting with 4290 = 6

Total=2×64+63+5×62+6+3=2997



Q 13 :

If Pn-1:Pn=11:21,2n-12n+1 then n2+n+15 is equal to ______ .                [2023]



(45)

Given, Pn-12n+1:2n-1Pn=11:21

(2n+1)!(2n+1-n+1)!×(n-1)!(2n-1)!=1121

(2n+1)!(n+2)!×(n-1)!(2n-1)!=1121

(2n+1)(2n)(2n-1)!(n+2)(n+1)(n)(n-1)!×(n-1)!(2n-1)!=1121

(2n+1)(2n)(n+2)(n+1)(n)=11214n+2n2+3n+2=1121

84n+42=11n2+33n+22 11n2-51n-20=0

(n-5)(11n+4)=0 n=5, -411

The value of n can never be negative.  

So, n=5

   n2+n+15=(5)2+5+15=45



Q 14 :

If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at 315th position in this arrangement is                [2024]

  • NRAPGU

     

  • NRAGPU

     

  • NRAPUG

     

  • NRAGUP

     

(1)

Letter in the word NAGPUR = 6

If we fix A _ _ _ _ _

Number of words start with A = 5! = 120

If we fix G _ _ _ _ _

Number of words start with G = 5! = 120

If we fix N A _ _ _ _

Number of words start with NA = 4! = 24

If we fix N G _ _ _ _

Number of words start with NG = 4! = 24

If we fix N P _ _ _ _

Number of words start with NP = 4! = 24

Next word i.e., 313th word in dictionary will be NRAGPU314th word will be NRAGUP315th word will be NRAPGU.



Q 15 :

60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50th word is                      [2024]

  • OBBJH

     

  • HBBJO

     

  • OBBHJ

     

  • JBBOH

     

(1)

Let us fix B first (Dictionary Order)

B _ _ _ _    (We have 4 distinct letters left, to be arranged in 4 distinct ways)
Number of ways = 4! = 24

H _ _ _ _   (We have B, B, J, O to be arranged in 4 ways when H is fixed)

Number of ways = 4!2!=12

J _ _ _ _    (B, B, H, O to be arranged in 4 ways)

Number of ways = 4!2!=12

49th word will be OBBHJ
50th word will be OBBJH

 



Q 16 :

If n is the number of ways in which five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to                     [2024]

  • 43

     

  • 47

     

  • 53

     

  • 51

     

(4)

Number of ways in which five different employees can sit are

I              II              III             IV

                                 

5             0               0                0             5!/5! = 1

4             1               0                0             5!/4! = 5

3             2               0                0            5!/3!2! = 10

3             1               1                0             5!/3! 1! 1! 2! = 10

2             2               1                0             5!2!2!1!2!=15

2            1                1                1              5!2!1!1!1!3!=10

        Total number of ways =1 + 5 + 10 + 10 + 15 + 10 = 51

 



Q 17 :

Let α=(4!)!(4!)3! and β=(5!)!(5!)4!. Then

  • αN and βN

     

  • αN and βN

     

  • αN and βN

     

  • αN and βN

     

(3)

 



Q 18 :

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to             [2024]

  • 18

     

  • 15

     

  • 12

     

  • 16

     

(2)

Ways of arranging 8 identical books in 4 identical shelves = {8, 0, 0, 0}, {7, 1, 0, 0}, {6, 2, 0, 0}, {5, 3, 0, 0}, {4, 4, 0, 0}, {6, 1, 1, 0},{5, 2, 1, 0}, {4, 3, 1, 0}, {4, 2, 2, 0}, {3, 3, 2, 0}, {5, 1, 1, 1}, {4, 2, 1, 1}, {3, 3, 1, 1}, {3, 2, 2, 1}, {2, 2, 2, 2}.

Number of ways = 15

[As shelves and books are identical, so the arrangements (5, 2, 1, 0), (2, 5, 1, 0), and (1, 2, 5, 0) are considered as same]

 



Q 19 :

The number of ways of getting a sum 16 on throwing a dice four times is ______.                 [2024]



(125)

We need sum of 16 on throwing a dice four times so possible ways are:
{(4, 4, 4, 4), (5, 4, 4, 3), (5, 5, 1, 5), (5, 5, 3, 3), (5, 5, 4, 2), (6, 4, 3, 3), (6, 4, 4, 2), (6, 5, 3, 2), (6, 5, 4, 1), (6, 6, 3, 1), (6, 6, 2, 2)}

Obtained Result Number of Ways
(6, 6, 2, 2) 4!2!2!=6
(6, 6, 3, 1) 4!2!=12
(6, 5, 4, 1) 4!=24
(6, 5, 3, 2) 4!=24
(6, 4, 4, 2) 4!2!=12
(6, 4, 3, 3) 4!2!=12
(5, 5, 4, 2) 4!2!=12
(5, 5, 3, 3) 4!2!2!=6
(5, 5, 1, 5) 4!3!=4
(5, 4, 4, 3) 4!2!=12
(4, 4, 4, 4) 1!=1
Total ways 125

 



Q 20 :

The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______.           [2024]



(70)

Number between 100 to 1000 are 3-digit number.

Let number abc such that a+b+c=14

where a,b,c{0,1,2,...9} and a1

Case I : All three digits are same ie., a=b=c

3a=14, which is not possible

Case II : Two digits are same i.e., a=bc

2a+c=14

  (a, c) = {(3, 8), (4, 6), (5, 4), (6, 2), (7, 0)}

In each subcase, number of ways of forming a 3-digit number = 3!2!

There are 5 such cases as (a, c) have 5 elements in which 0,7,7 is included

So, total number of 3 digits numbers when 2 digits are same 5×3!2!-1

=15-1=14


Case III : All digits are different

   (a, b, c) = {(1, 4, 9), (2, 4, 8), (2, 3, 9), (1, 5, 8), (3, 4, 7), (2, 5, 7), (1, 6, 7), (3, 5, 6), (5, 9, 0), (6, 8, 0)}


Total number of 3 digits number formed by above triplet = 10×3!-2×2!=60-4=56


Total number of required number = 56 + 14 = 70



Q 21 :

All the letters of the word “GTWENTY” are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word “GTWENTY” is _____.             [2024]



(553)

Number of words starts with E=6!2!

Number of words starts with GE=5!2!

Number of words starts with GN=5!2!

Number of words starts with GTE=4!

Number of words starts with GTN=4!

Number of words starts with GTT=4!

Next word will be GTWENTY

     Rank of GTWENTY =6!2!+2×5!2!+3×4!+1=553

 



Q 22 :

The largest nN such that 3n divides 50! is :          [2025]

  • 23

     

  • 22

     

  • 21

     

  • 20

     

(2)

The number of times a prime p divides n! is k=1[npk]

Required largest value of

n=[503]+[5032]+[5033]

   = 16 + 5 + 1 = 22

Hence, the maximum value of n is 22.



Q 23 :

The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :          [2025]

  • 360

     

  • 1820

     

  • 45

     

  • 2520

     

(4)

Given number in the sequences are 11111 222 00

  Number of sequences = 10!5!3!2! = 2520



Q 24 :

In a group of 3 girls and 4 boys, there are two boys B1 and B2. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but B1 and B2 are not adjacent to each other, is :          [2025]

  • 144

     

  • 120

     

  • 96

     

  • 72

     

(1)

Total number of ways

=3!×4!×2!3!×3!×2!×2!=144



Q 25 :

The Number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is          [2025]

  • 34000

     

  • 36000

     

  • 37000

     

  • 35000

     

(2)

Number of words in which vowels never come together

= Total number of words – Number of words in which vowels come together

=8!6!×3! = 36000



Q 26 :

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is          [2025]

  • 4608

     

  • 4607

     

  • 5719

     

  • 5720

     

(2)

Given digits are 0, 1, 2, 3, 4, 5, 6, 7.

First place can be filled in 3ways, i.e., (5, 6, 7)

For 1st place as 5, last place can be 0, 1, 2, 3

   Total number of ways = 1×512×4 = 2048

For 1st place as 6, last place can be 0, 1, 2

   Total number of ways = 1×512×3 = 1536

For 1st place as 7, last place can be 0, 1

Total number of ways = 1×512×2 = 1024

   Total number of ways = 2048 + 1536 + 1024 = 4608

So, 50000 is not included i.e., 4608 – 1 = 4607



Q 27 :

Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is :          [2025]

  • 158

     

  • 164

     

  • 173

     

  • 161

     

(4)

(i) Number of numbers formed using 1 and 3, i.e.,

     =7!5!2!=21

(ii) Number of numbers formed using 1, 2, 3 i.e.,

       =7!4!2!=105

(iii) Number of numbers formed using 1 and 2, i.e.,

       =7!3!4!=35

Number of elements in set P = 161.



Q 28 :

If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement, is :          [2025]

  • PRNAUK

     

  • PRKANU

     

  • PRKAUN

     

  • PRNAKU

     

(3)

We have, A, K, N, P, R, U

Now, to calculate rank

Number of words start with A               = 5! = 120

Number of words starts with K             = 5! = 120

Number of words starts with N             = 5! = 120

Number of words starts with PA           = 4! =   24

Number of words starts with PK           = 4! =   24

Number of words starts with PN           = 4! =   24

Number of words starts with PRA         = 3! =     6

Number of words starts with PRKANU = 1

Number of words starts with PRKAUN = 1

                                                  Total = 440

  440th words is PRKAUN.



Q 29 :

If the number of seven-digit numbers, such that the sum of their digits is even, is m·n·10n; m, n{1,2,3,...,9}, then m + n is equal to __________.          [2025]



(14)

Total 7 digit numbers = 9000000

7 digit numbers having sum of their digits is even

        =90000002=9·5·105

On comparing with m·n·10n, we get

         m = 9 and n = 5

    m + n = 14.



Q 30 :

Let m and n, (m < n), be two 2-digit numbers. Then the total numbers of pairs (m, n), such that gcd (m, n) = 6, is __________.          [2025]



(64)

Given m and n are 2-digit numbers (m < n).

Such that gcd (m, n) = 6.

Let m = 6a, n = 6b, where a and b are coprime integers a < b.

 10m99, 10n99.

 2a16, 2b16.

Now, if a = 2, then b = 3, 5, 7, 9, 11, 13, 15 = 7

a = 3, then b = 4, 5, 7, 8, 10, 11, 13, 14, 16 = 9

a = 4, then b = 5, 7, 9, 11, 13, 15 = 6

a = 5, then b = 6, 7, 8, 9, 11, 12, 13, 14, 16 = 9

a = 6, then b = 7, 11, 13 = 3

a = 7, then b = 8, 9, 10, 11, 12, 13, 15, 16 = 8

a = 8, then b = 9, 11, 13, 15 = 4

a = 9, then b = 10, 11, 13, 14, 16 = 5

a = 10, then b = 11, 13 = 2

a = 11, then b = 12, 13, 14, 15, 16 = 5

a = 12, then b = 13 = 1

a = 13, then b  = 14, 15, 16 = 3

a = 14, then b = 15 = 1

a = 15, then b = 16 = 1

  Total possible number of ordered pairs = 64.



Q 31 :

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is __________.          [2025]



(17280)

Number of ways that all boys sit together =5!×5!

Number of ways no two boys sit together =4!×5!

   Required number of ways =5!×5!+4!×5!=17280



Q 32 :

The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is __________.          [2025]



(125)

Number of 3-digits = 999 – 99 = 900

Number of 3-digit numbers divisible by 2 & 3 i.e., by 6,

        9006=150

Number of 3-digit numbers divisible by 4 & 9 i.e., by 36,

         90036=25

  Number of 3-digit numbers divisible by 2 & 3 but not 4 & 9 = 150 – 25 = 125.



Q 33 :

Number of functions f:{1,2,...,100}{0,1}, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to __________.          [2025]



(392)

Given : f:{1,2,...,100}{0,1}

Number of ways to connect {1, 2, ..., 98} to 1 = 98

Number 99 can connect either 0 or 1  2 ways

Similarly, 100 can connect either 0 or 1  2 ways

   Total number of functions for the given condition that assign 1 to exactly one of positive integers  98 is given by 98×2×2 = 392.



Q 34 :

The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is __________.          [2025]



(64)

Let xyz be any number between 212 and 999

Let x=2y+z=13, then

(y, z) : (4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4), i.e., 6 in number.

Let x=3y+z=12, then

(y, z) : (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3) i.e., 7 in number.

Let x=4y+z=11, then

(y, z) : (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4),(8, 3), (9, 2) i.e., 8 in number.

Let x=5y+z=10, then

(y, z) : (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3). (8, 2), (9, 1) i.e., 9 in number.

Let x=6y+z=9, then

(y, z) : (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0) i.e., 10 in number.

Let x=7y+z=8, then

(y, z) : (0, 8), (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0) i.e., 9 in number.

Let x=8y+z=7, then

(y, z) : (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6,1), (7, 0) i.e., 8 in number.

Let x=9y+z=6, then

(y, z) : (0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0) i.e., 7 in number.

Total = 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64.



Q 35 :

The number of 4-letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is __________.           [2026]



(1422)

P3, Q2, R2, S,T,U,V

Case I: 3 alike, 1 different

C11×C16×4!3!=24

Case II: 2 alike, 2 alike

C23×4!2!2!=18

Case III: 2 alike, 2 different

C13×C26×4!2!=540

Case IV: All 4 different

C47×4!=840

Total words=1422



Q 36 :

The number of relations, defined on the set {a, b, c, d}, which are both reflexive and symmetric, is equal to:        [2026]

  • 64

     

  • 256

     

  • 16

     

  • 1024

     

(1)

Number of relation which are reply and sym. both =14×26=64

(a, a) (a, b) (a, c) (a, d)
(b, a) (b, b) (b, c) (b, d)
(c, a) (c, b) (c, c) (c, d)
(d, a) (d, b) (d, c) (d, d)



Q 37 :

The number of strictly increasing functions f from the set {1,2,3,4,5,6} to the set {1,2,3,,9} such that f(i)i for 1i6, is equal to   [2026]

  • 28

     

  • 22

     

  • 27

     

  • 21

     

(1)

f(i)i,  f(x) is strictly increasing function  

f:AB,  where A={1,2,3,,6}

B={1,2,3,,9}, then number of functions f:AB is equal to

f(i)i   Case (i)  f(1)=2   C57=21

                 Case (ii)  f(1)=3   C56=6

                  Case (iii)  f(1)=4   C55=1

Number of functions from A to B =21+6+1=28



Q 38 :

The number of 3×2 matrices A, which can be formed using the elements of the set {-2,-1,0,1,2} such that the sum of all the diagonal elements of ATA is 5, is ________ .                           [2026]



(312)

(a1b1a2b2a3b3)3×2

ATA=(a1a2a3b1b2b3)2×3(a1b1a2b2a3b3)3×2

=(a12+a22+a32b12+b22+b32)

Tr(ATA)=a12+a22+a32+b12+b22+b32=5

{2,1,0,0,0,0}

{2,-1,0,0,0,0}

{-2,1,0,0,0,0}

{-2,-1,0,0,0,0}

{1,1,1,1,1,0}

No. of ways=6!4!×4+2×6!5!+2×6!4!+2×6!3!2!

=6!3!+2×6×2+2×15×2×6!3!

=120+120+12+60=312



Q 39 :

The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______.        [2026]



(42)

(1) all different

  5,0,1,9C34=6 ways

(2) 2 alike, 2 different

  5,0,0,13 ways

  5,1,1,23 ways

  5,2,2,03 ways

  5,2,2,93 ways

  5,5,0,26 ways

  5,5,2,96 ways

  5,1,9,93 ways

(3) 3 alike, 1 different

  5,5,5,03 ways

  5,5,5,93 ways

(4) 2 alike, 2 other alike

  5,5,1,13 ways

Total ways=42



Q 40 :

Three persons enter in a lift at the ground floor. The lift will go upto 10th floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ________.  [2026]



(210)

C37×3!=7×6×51×2×3×3!=210