Topic Question Set


Q 41 :    

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 42 :    

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 43 :    

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 44 :    

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 45 :    

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 46 :    

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 47 :    

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 48 :    

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 49 :    

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 50 :    

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.