# CBSE Solutions for Class 11 English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Evaluate each of the following : 3P3

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

Evaluate each of the following : 9P1

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

Evaluate each of the following : P (10, 2)

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

Evaluate each of the following : 5P5

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

Evaluate each of the following : 10P1

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

Evaluate each of the following : 10P2

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

Evaluate each of the following : P (6, 2)

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

Evaluate each of the following : 6P5

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

Evaluate each of the following : 10P3

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

Evaluate each of the following : 8P2

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

If 10C0 = 10Cr, find r

જવાબ : 10

If 40C20 = 20Cr, find r

જવાબ : 20

If 10C4 = 20Cr, find r

જવાબ : 6

If nC4 = nC6, find n

જવાબ : 10

If nC4 = nC3, find n

જવાબ : 7

If nC1 = nC6, find n

જવાબ : 7

If 20C4 = 20Cr, find r

જવાબ : 16

If nC4 = nC46, find n

જવાબ : 50

If nC4 = nC60, find n

જવાબ : 64

20C3 = 20Cr, find r

જવાબ : 17

If 8Cr − 7C1 = 7C0, find r.

જવાબ : 1

If 99Cr − 98C3 = 98C2, find r.

જવાબ : 3

If 12Cr − 11C10 = 11C9, find r.

જવાબ : 10

If 6Cr − 5C3 = 7C2, find r.

જવાબ : 3

If 8Cr − 7C6 = 7C5, find r.

જવાબ : 6

If 8Cr − 7C2 = 7C1, find r.

જવાબ : 2

If 45C5 = 45Cr, find r

જવાબ : 40

If 27C7 = 27Cr, find r

જવાબ : 20

If 24C4 = 24Cr, find r

જવાબ : 20

If 18C4 = 18Cr, find r

જવાબ : 14

Evaluate each of the following : 7C7

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

Evaluate each of the following : 9C2

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

Evaluate each of the following : C (10, 5)

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

Evaluate each of the following : 5C3

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

Evaluate each of the following : 10C8

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

Evaluate each of the following : 10C9

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

Evaluate each of the following : C (6, 3)

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

Evaluate each of the following : 8C2

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

Evaluate each of the following : 6C5

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

Evaluate each of the following : C (6, 2)

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

Evaluate each of the following : 10C2

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

Evaluate each of the following :  10C1

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

Evaluate each of the following : 5C5

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

Evaluate each of the following : C (10, 2)

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

Evaluate each of the following : 9C1

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

Evaluate each of the following : 3C3

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

Evaluate each of the following : 8C3

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

Evaluate each of the following : 10C2

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

Evaluate each of the following : 6C6

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

Evaluate each of the following : 10C3

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

A sports team of 11 students is to be constituted, choosing at least 5 from class IX and at least 5 from class X. If there are 20 students in each of these classes, in how many ways can the teams be constituted?

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)

From 4 officers and 8 militants in how many ways can 6 be chosen to include exactly one officer?

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

How many different selections of 4 notebook can be made from 10 different notebooks, if there is no restriction

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

From a class of 12 girls and 10 boys, 10 students are to be chosen for a competition; at least including 4 girls  and 4 boys. The 2 boys who won the prizes last year should be included. In how many ways can the selection be made?

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

How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 13 (without repetition)?

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

There are 10 teachers and 20 college students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done.

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

In how many ways can a cricket team of 11 players be selected from 16 players?

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

In how many ways can a school student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?

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

How many different boat parties of 8, consisting of 3 boys and 5 girls, can be made from 10 boys and 25 girls?

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

From a group of 13 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?

જવાબ : Required number of ways = 13C11

Now,13C11 =13C2
= 13/2 × 12/1 × 11C0

=78

There are 5 books on Mathematics and 6 books on Chemistry  in a book shop. In how many ways can a students buy a Mathematics book and a Chemistry

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

By using fundamental principle of multiplication:
Number of ways of buying a mathematics and a Chemistry book = 6×5 = 30

There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 2 each?

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

A letter lock consists of 2 rings each marked with 10 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock?

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

In how many ways can an examinee answer a set of 9 true/false type questions?

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

A coin is tossed 4 times and outcomes are recorded. How many possible outcomes are there?

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

There are three parcels and five post-offices. In how many different ways can the parcels be sent by registered post?

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

A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets (one plate for each month). How many types of calendars should it prepare to serve for all the possibilities in future years?

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

From Chennai to Bombay there are two roots; air, and sea. From Bombay to Delhi there are three routes; air, rail and road. From Chennai to Delhi via Bombay, how many kinds of routes are there?

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

A person wants to buy one Blue pen, one Black pen and one pencil from a stationery shop. If there are 10 Blue pen varieties, 12 Black pen varieties and 5 pencil varieties, in how many ways can he select these articles?

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

In a class there are 14 boys and 27 girls. The teacher wants to select 1 boy and 1 girl to represent the class in a function. In how many ways can the teacher make this selection?

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

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 1 8P2 A 30 2 10P3 B 6 3 6P1 C 720 4 P(6,2) D 56

જવાબ :

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

 1 10P2 A 120 2 10P1 B 9 3 5P5 C 90 4 P(9,1) D 10

જવાબ :

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

 1 9P1 A 7 2 3P1 B 9 3 6P1 C 3 4 7P1 D 6

જવાબ :

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

Find r

 1 10C0 = 10Cr A 14 2 18C4 = 18Cr B 20 3 24C4 = 24Cr C 10

જવાબ :

1-C, 2-A, 3-B

Find n

 1 nC4 = nC6 A 50 2 nC4 = nC3 B 64 3 nC4 = nC60 C 10 4 nC4 = nC46 D 7

જવાબ :

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

Find r

 1 20C4 = 20Cr A 20 2 20C3 = 20Cr B 6 3 10C4 = 20Cr C 17 4 40C20 = 20Cr D 16

જવાબ :

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

 1 27C7 = 27Cr A 40 2 45C5 = 45Cr B 2 3 8Cr − 7C2 = 7C1 C 20

જવાબ :

1-C, 2-A, 3-B

 1 8Cr − 7C6 = 7C5 A 3 2 6Cr − 5C3 = 7C2 B 10 3 12Cr − 11C10 = 11C9 C 1 4 8Cr − 7C1 = 7C0 D 6

જવાબ :

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

 1 8C2 A 6 2 10C3 B 120 3 6C5 C 28

જવાબ :

1-C, 2-B, 3-A

 1 C(6,2) A 10 2 10C2 B 15 3 10C1 C 45

જવાબ :

1-B,2-C,3-A