Topic Question Set


Q 11 :    

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 12 :    

The umber of words, which can be formed using all the letters of the word "DOUGHTER", 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 whichvowels come together

=8!6!×3! = 36000



Q 13 :    

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  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 14 :    

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 15 :    

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 16 :    

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 17 :    

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 18 :    

The number of the 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 19 :    

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 20 :    

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 exactly one of positive integers  98 is given 98×2×2 = 392.