CBSE Solutions for Class 11 English

જવાબ : ³P₃ = 3! / (3−3)!   = 3! /0! = 3!  (Since, 0! = 1)            = 6

જવાબ : ⁹P₁ =9! / (9−1)!   = 9! /8!   =9

જવાબ : P (10, 2)
it can also be written as ¹⁰P_2.
¹⁰P₂ = 10! /8!         = 10(9)      = 90

જવાબ : ⁵P₅ = 5! / (5−5)!  = 5! /0!  = 5! (Since, 0! = 1)  = 120

જવાબ : ¹⁰P₁ =10! / (10−1)!    = 10! /9!  =10

જવાબ :  ¹⁰P₂ =10! / (10−2)! = 10! /8! = 9x10 = 90

જવાબ : P (6, 2)
it can also be written as ⁶P_2.
⁶P₂ = 6! /4!         = 6(5)      = 30

જવાબ : ⁶P₅ = 6! / (6−5)!= 6! /1! = 6!(Since , 0! = 1)  = 720

જવાબ : ¹⁰P₃ =10! / (10−3)!= 10! /7!=10(9) (8) (7!)/7! =10×9×8  = 720 

જવાબ :  ⁸P₂ =8! / (8−2)! = 8! /6! =8(7) (6!)/ 6! =8×7 = 56 

જવાબ : 10

જવાબ : 20

જવાબ : 6

જવાબ : 10

જવાબ : 7

જવાબ : 7

જવાબ : 16

જવાબ : 50

જવાબ : 64

જવાબ : 17

જવાબ : 1

જવાબ : 3

જવાબ : 10

જવાબ : 3

જવાબ : 6

જવાબ : 2

જવાબ : 40

જવાબ : 20

જવાબ : 20

જવાબ : 14

જવાબ : ⁷C₇ = 7! / (7−7)! 7!   = 7! /7! = 1

જવાબ : ⁹C₁ =9! / (9−2)! 2!   = 9! /7! ×2  =36

જવાબ : C (10, 5)
It can also be written as ¹⁰C_5.
¹⁰C₅ = 10! /5! 5!         = 10×9×8×7×6/5×4×3×2      = 252

જવાબ : ⁵C₅ = 5! / (5−3)! ×3!             = 10

જવાબ : ¹⁰C₁ =10! / (10−8)! ×8!= 10! /8! ×2=5

જવાબ :  ¹⁰C₁ =10! / (10−9)! ×9! = 10! /9! ×21 = 10

જવાબ : C (6, 3)
It can also be written as ⁶C_3.
⁶C₃ = 6! /3! ×3!         = 6×5×4/6      = 20

જવાબ :  ⁸C₂ =8! /2! (8−2)! = 8! /2! 6! =8(7) (6!)/2×6! =4×7 = 28

જવાબ : ⁶C₅ = 6! / (6−5)! 5!= 6! /1! ×5!   = 6

જવાબ : C (6, 2)
It can also be written as ⁶C_2.
⁶C₂ = 6! /4! ×2         = 6(5)/2      = 15

જવાબ :  ¹⁰C₂ =10! / (10−2)! ×2 = 10! /8! ×2 = 9x10/2 = 45

જવાબ : ¹⁰C₁ =10! / (10−1)! × 1             = 10! /9!             =10

જવાબ :  ⁵C₅ = 5! / (5−5)! ×5!             = 1

જવાબ : C (10, 2)
It can also be written as ¹⁰C_2.
¹⁰C₂ = 10! /8! 2!         = 10(9)/2      = 45

જવાબ : ⁹C₁ =9! / (9−1)! 1!              = 9! /8!             =9

જવાબ : 3C₃ = 3! / (3−3)! 3!             = 3! /3!       = 1

જવાબ :  ⁸C₂ =8! /3! (8−3)! = 8! /3! 5! =8(7) (6) (5!)/3! ×5! =8×7×6/6 = 56

જવાબ : ¹⁰C₂ =10! / (10−2)! 2!              = 10! /8! ×2!             =10(9) (8!)/8! ×2            =45

જવાબ : ⁶C₅ = 6! / (6−6)! 6! = 6! /0! ×6! = 1

જવાબ : ¹⁰C₃ =10! / (10−3)! 3!= 10! /7! ×3! =10(9) (8) (7!)/7! ×6  =10×9×8/6 = 120 

જવાબ : A sports team of 11 students is to be constituted, choosing at least 5 students of class IX and at least 5 from class X.
Required number of ways = ²⁰C₅ × ²⁰C₆ + ²⁰C₆ × ²⁰C₅  = 2 ×²⁰C₅ × ²⁰C₆ = 2 (²⁰C₆ × ²⁰C₅)

જવાબ : From 4 officers and 8 millitants, 6 need to be chosen. Out of them, 1 is an officer.
   Required number of ways = ⁴C₁ × ⁸C₅  = 4 × 8!/5! 3! = 4 × 8 × 7 × 6 / 6 = 224

જવાબ : Required ways of selecting 4 notebooks from 10 notebooks without any restriction = ¹⁰C₄ = 10/4 × 9/3 × 8/2 × 7 = 210

જવાબ : Two girls who won the prizes last year are to be included in every selection.
So, we have to select 8 students out of 12 girls and 8 boys, choosing at least 4 girls and 2 boys.
Number of ways in which it can be done = ¹²C₆ × ⁸C₂  +  ¹²C₅ × ⁸C₃   + ¹²C₄ × ⁸C₄  =  25872 + 44352 + 34650 = 104874

જવાબ : Required number of ways of getting different products = ⁴C₂ + ⁴C₃  + ⁴C₄   = 6 + 4 + 1 = 11

જવાબ : Clearly, 2 teachers and 3 students are selected out of 10 teachers and 20 students, respectively.
    Required number of ways  = ¹⁰C₂ × ²⁰C₃  = 10/2 × 9/1 × 20/3 × 19/2 × 18/1  = 51300

જવાબ : Number of ways in which 11 players can be selected out of 16 = ¹⁶C₁₁ = 16!/11! 5! = 16×15×14×13×12/5×4×3×2×1 = 4368

જવાબ : We are given that 2 courses are compulsory out of the 9 available courses,
Thus, a student can choose 3 courses out of the remaining 7 courses.
Number of ways = ⁷C₃ = 7!/3! 4! = (7×6×5)/(3×2×1) = 35

જવાબ : Clearly, out of the 10 boys and 25 girls, 3 boys and 5 girls will be chosen.

Then, different boat parties of 8 = ²⁵C₅ × ¹⁰C₃                                    =25!/(5! 20!) ×10!/(3! 7!) = (25×24×23×22×21)/(5×4×3×2×1) × (10×9×8)/(3×2×1) = 6375600

જવાબ : Required number of ways = ¹³C₁₁
 
Now,¹³C₁₁ =¹³C₂                                                    
            = 13/2 × 12/1 × 11C0                
  
             =78

જવાબ : Number of  books on mathematics = 5
Number of books on Chemistry  = 6
Number of ways of buying a mathematics book = 5
Similarly, number of ways of buying a Chemistry book = 6

જવાબ : Number of ways of answering the first three questions = 4 each
Number of ways of answering the remaining three questions = 2 each
∴ Total number of ways of answering all the questions = 4×4×4×2×2 = 256 

જવાબ : Number of ways of marking each of the ring = 10 different letters
∴ Total number of ways of marking any letter on these three rings = 10×10 = 100
Out of these 100 combinations of the lock, 1 combination will be successful.
∴ Total number of unsuccessful attempts = 100 − 1 = 99

જવાબ : Number of ways of answering the first question = 2 (either true or false)
Similarly, each question can be answered in 2 ways.
∴ Total number of ways of answering all the 9 questions = 2×2×2×2×2×2×2×2×2= 2⁹ = 512

જવાબ : Number of outcomes when the coin is tossed for the first time = 2
Number of outcomes when the coin is tossed for the second time = 2
Thus, there would be 2 outcomes, each time the coin is tossed.
Total number of possible outcomes on tossing the coin five times = 2×2×2×2 = 16

જવાબ : Number of ways of sending 1 parcel via registered post = 5
Number of ways of sending 3 parcels via registered post through 5 post offices = 5×5×5 = 125

જવાબ : The first day of the year can be any one of the days of the week, i.e the first day can be selected in 7 ways.
But, the year could also be a leap year.
So, the mint should prepare 7 calendars for the non-leap year and 7 calendars for the leap year.
So, total number of calendars that should be made = 7 + 7 = 14

જવાબ : Number of routes from Chennai to Bombay = 2
Number of routes from Bombay to Delhi = 3
Using fundamental principle of multiplication:
Number of routes from Chennai to Delhi via Bombay = 2 × 3 = 6

જવાબ : Number of blue  pen varieties = 10
Number of  black pen varieties = 12
Number of pencil varieties = 5
Ways to select a blue pen = 10
Ways to select a black pen = 12
Ways to select a pencil = 5

Ways to select a blue pen, a black pen and a pencil = 10 × 12 × 5 = 600 

જવાબ : No. of boys in the class = 14
No. of girls in the class = 27
Ways to select a boy = 14
Similarly, ways to select a girl = 27
∴ Number of ways to select 1 boy and 1 girl = 14 × 27 = 378

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-C, 2-D, 3-A, 4-B

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-C, 2-A, 3-B

જવાબ :

1-C, 2-D, 3-B, 4-A

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-C, 2-A, 3-B

જવાબ :

1-D, 2-A, 3-B, 4-C

જવાબ :

1-C, 2-B, 3-A

જવાબ :

1-B,2-C,3-A