Topic Question Set


Q 1 :    

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

     

(A)

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 2 :    

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

     

(A)

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 3 :    

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

     

(D)

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 4 :    

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

  • αN and βN

     

  • αN and βN

     

  • αN and βN

     

  • αN and βN

     

(C)

 



Q 5 :    

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

     

(B)

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 6 :    

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 7 :    

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 8 :    

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