Topic Question Set


Q 11 :    

The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition, is       [2023]

  • 168

     

  • 120 

     

  • 220

     

  • 48

     

(1)

5 digits numbers = 5! = 120

Four digit number greater than 7000 = 2 × 4 × 3 × 2 = 48

Total number greater than 7000 = 120 + 48 = 168

 



Q 12 :    

The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is            [2023]

  • 72 

     

  • 120

     

  • 6

     

  • 12

     

(1)

Number of numbers lies between 5000 and 10000 are 4 digit numbers greater than 5000 so, the first digit can be filled in 3 ways (using 5, 7, 9) and as repetition is not allowed, the other choices are 4, 3, 2.

Total number of number = 3 × 4 × 3 × 2 = 72

 



Q 13 :    

The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is       [2023]

  • 86

     

  • 89 

     

  • 79

     

  • 84

     

(2)

The letters in sequence of the word ‘OUGHT’ in alphabetical order as follows: G, H, O, T, U
Word starts with G can arrange in 4! ways.
                           H can arrange in 4! ways.
                           O can arrange in 4! ways.
                           TG can arrange in 3! ways.
                           TH can arrange in 3! ways.
                           TOG can arrange in 2! ways.
                           TOH can arrange in 2! ways.
                           TOUGH can arrange in 1! way.

So we must have serial number of the word is 89.

 



Q 14 :    

The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is           [2023]

  • 400

     

  • 507

     

  • 432

     

  • 472

     

(3)

We know that we have total 3-digit numbers only 900.

   Total 3 digit numbers=900

Out of these 900, the number divisible by 3 =300     (9003=300)

In the same way, the numbers divisible by 4 =9004=225

So, divisible by 3 and 4 = 75 (Using 90012=75)

Number divisible by either 3 or 4 =300+225-75=450

Now, we have to remove the numbers divisible by 48, 144, 192 ....... 18 terms.

Hence, required numbers, who are divisible by either 3 or 4 but not by 48 is 450-18=432



Q 15 :    

The number of ways of selecting two numbers a and b,a{2,4,6...,100} and b{1,3,5...,99} such that 2 is the remainder when a+b is divided by 23 is    [2023]

  • 54 

     

  • 108

     

  • 268

     

  • 186

     

(2)

a{2,4,6,,100},  b{1,3,5,,99}

a+b=23λ+2 

λ=0,1,2,,22 but λ can't be even. 

If λ=1,(a,b)12 pairs;  λ=3,(a,b)35 pairs; 

   λ=5,(a,b)42 pairs;  λ=7,(a,b)19 pairs; 

   λ=9,(a,b)0 pairs. 

   Total number of ways=12+35+42+19=108



Q 16 :    

The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ________ .         [2023]



(4898)

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

Total number of permutations = 7!

Let A = number of numbers containing string 153

Let B = number of numbers containing string 2467

n(A)=5!×1, n(B)=4!×1, n(AB)=2!

Required number of permutations = Total number of permutations - n(AB)

=7!-(5!+4!-2!)=7!-142=4898



Q 17 :    

The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to _________ .          [2023]



(26664)

Sum of all unit place numbers

=3!×2+3!2!×3+3!2!×1=12+9+3=24

   Sum of all 4 digit numbers:

=1×24+10×24+100×24+1000×24

=24000+2400+240+24=26,664



Q 18 :    

In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is __________ .          [2023]



(44)

Dearrangement of 5 students

D5=5!(1-11!+12!-13!+14!-15!)

=120(12-16+124-1120)=60-20+5-1=44

 



Q 19 :    

The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is __________ .     [2023]



(413)

Case 1: 1111143

Number of ways = 7!5!=42                    [Since, 1 repeats five times]

Case 2: 1111332

Number of ways = 7!4!2!=105             [Since, 1 repeats 4 times and 3 repeats 2 times]

Case 3: 2223111

Number of ways =  7!3!3!=140            [Since, 2 repeats 3 times and 1 repeats 3 times]

Case 4: 2222211

Number of ways = 7!5!2!=21                [Since, 2 repeats 5 times and 1 repeats 2 times]

Case 5: 1111422

Number of ways = 7!4!2!=105              [Since, 1 repeats 4 times and 2 repeats 2 times]

   Total number of ways=42+105+140+21+105=413



Q 20 :    

Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is ________ .         [2023]



(16)

Total 3-digit numbers formed by 1, 2, 3, 4, 5 when the last digit in 4 and divisible by 6.

[IMAGE 18]

 

111441142!232!352!]=8

When last digit is 2:

[IMAGE 19]

221551132!252!432!]=8

Total 3-digit numbers formed by 1, 2, 3, 4, 5 which are divisible by 6 are 16.