LOADING . . .
જવાબ : ^{3}P_{3 }= 3! / (3−3)! = 3! /0! = 3! (Since, 0! = 1) = 6
જવાબ : ^{9}P_{1 }=9! / (9−1)! = 9! /8! =9
જવાબ : P (10, 2)
it can also be written as ^{10}P_{2.}
^{10}P_{2} = 10! /8! = 10(9) = 90
જવાબ : ^{5}P_{5 }= 5! / (5−5)! = 5! /0! = 5! (Since, 0! = 1) = 120
જવાબ : ^{10}P_{1 }=10! / (10−1)! = 10! /9! =10
જવાબ : ^{10}P_{2} =10! / (10−2)! = 10! /8! = 9x10 = 90
જવાબ : P (6, 2)
it can also be written as ^{6}P_{2.}
^{6}P_{2} = 6! /4! = 6(5) = 30
જવાબ : ^{6}P_{5 }= 6! / (6−5)!= 6! /1! = 6!(Since , 0! = 1) = 720
જવાબ : ^{10}P_{3 }=10! / (10−3)!= 10! /7!=10(9) (8) (7!)/7! =10×9×8 = 720
જવાબ : ^{8}P_{2} =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 : ^{7}C_{7 }= 7! / (7−7)! 7! = 7! /7! = 1
જવાબ : ^{9}C_{1 }=9! / (9−2)! 2! = 9! /7! ×2 =36
જવાબ : C (10, 5)
It can also be written as ^{10}C_{5.}
^{10}C_{5} = 10! /5! 5! = 10×9×8×7×6/5×4×3×2 = 252
જવાબ : ^{5}C_{5 }= 5! / (5−3)! ×3! = 10
જવાબ : ^{10}C_{1 }=10! / (10−8)! ×8!= 10! /8! ×2=5
જવાબ : ^{10}C_{1} =10! / (10−9)! ×9! = 10! /9! ×21 = 10
જવાબ : C (6, 3)
It can also be written as ^{6}C_{3.}
^{6}C_{3} = 6! /3! ×3! = 6×5×4/6 = 20
જવાબ : ^{8}C_{2} =8! /2! (8−2)! = 8! /2! 6! =8(7) (6!)/2×6! =4×7 = 28
જવાબ : ^{6}C_{5 }= 6! / (6−5)! 5!= 6! /1! ×5! = 6
જવાબ : C (6, 2)
It can also be written as ^{6}C_{2.}
^{6}C_{2} = 6! /4! ×2 = 6(5)/2 = 15
જવાબ : ^{10}C_{2} =10! / (10−2)! ×2 = 10! /8! ×2 = 9x10/2 = 45
જવાબ : ^{10}C_{1 }=10! / (10−1)! × 1 = 10! /9! =10
જવાબ : ^{5}C_{5 }= 5! / (5−5)! ×5! = 1
જવાબ : C (10, 2)
It can also be written as ^{10}C_{2.}
^{10}C_{2} = 10! /8! 2! = 10(9)/2 = 45
જવાબ : ^{9}C_{1 }=9! / (9−1)! 1! = 9! /8! =9
જવાબ : 3C_{3 }= 3! / (3−3)! 3! = 3! /3! = 1
જવાબ : ^{8}C_{2} =8! /3! (8−3)! = 8! /3! 5! =8(7) (6) (5!)/3! ×5! =8×7×6/6 = 56
જવાબ : ^{10}C_{2 }=10! / (10−2)! 2! = 10! /8! ×2! =10(9) (8!)/8! ×2 =45
જવાબ : ^{6}C_{5 }= 6! / (6−6)! 6! = 6! /0! ×6! = 1
જવાબ : ^{10}C_{3 }=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 = ^{20}C_{5 }× ^{20}C_{6} + ^{20}C_{6} × ^{20}C_{5} = 2 ×^{20}C_{5} × ^{20}C_{6} = 2 (^{20}C_{6} × ^{20}C_{5})
જવાબ : From 4 officers and 8 millitants, 6 need to be chosen. Out of them, 1 is an officer.
Required number of ways = ^{4}C_{1} × ^{8}C_{5} = 4 × 8!/5! 3! = 4 × 8 × 7 × 6 / 6 = 224
જવાબ : Required ways of selecting 4 notebooks from 10 notebooks without any restriction = ^{10}C_{4} = 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 = ^{12}C_{6} × ^{8}C_{2} + ^{12}C_{5} × ^{8}C_{3} + ^{12}C_{4} × ^{8}C_{4} = 25872 + 44352 + 34650 = 104874
જવાબ : Required number of ways of getting different products = ^{4}C_{2 }+ ^{4}C_{3} + ^{4}C_{4} = 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 = ^{10}C_{2} × ^{20}C_{3} = 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 = ^{16}C_{11} = 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 = ^{7}C_{3} = 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 = ^{25}C_{5 }× ^{10}C_{3} =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 = ^{13}C_{11}
Now,^{13}C_{11} =^{13}C_{2}
= 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= 2^{9} = 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
1 |
^{8}P_{2} |
A |
30 |
2 |
^{10}P_{3} |
B |
6 |
3 |
^{6}P_{1} |
C |
720 |
4 |
P(6,2) |
D |
56 |
જવાબ :
1-D, 2-C, 3-B, 4-A
1 |
^{10}P_{2} |
A |
120 |
2 |
^{10}P_{1} |
B |
9 |
3 |
^{5}P_{5} |
C |
90 |
4 |
P(9,1) |
D |
10 |
જવાબ :
1-C, 2-D, 3-A, 4-B
1 |
^{9}P_{1} |
A |
7 |
2 |
^{3}P_{1} |
B |
9 |
3 |
^{6}P_{1} |
C |
3 |
4 |
^{7}P_{1} |
D |
6 |
જવાબ :
1-B, 2-C, 3-D, 4-A
Find r
1 |
^{10}C_{0} = ^{10}C_{r} |
A |
14 |
2 |
^{18}C_{4} = ^{18}C_{r} |
B |
20 |
3 |
^{24}C_{4} = ^{24}C_{r} |
C |
10 |
જવાબ :
1-C, 2-A, 3-B
Find n
1 |
^{n}C_{4} = ^{n}C_{6} |
A |
50 |
2 |
^{n}C_{4} = ^{n}C_{3} |
B |
64 |
3 |
^{n}C_{4} = ^{n}C_{60} |
C |
10 |
4 |
^{n}C_{4} = ^{n}C_{46} |
D |
7 |
જવાબ :
1-C, 2-D, 3-B, 4-A
Find r
1 |
^{20}C_{4} = ^{20}C_{r} |
A |
20 |
2 |
^{20}C_{3} = ^{20}C_{r} |
B |
6 |
3 |
^{10}C_{4} = ^{20}C_{r} |
C |
17 |
4 |
^{40}C_{20} = ^{20}C_{r} |
D |
16 |
જવાબ :
1-D, 2-C, 3-B, 4-A
1 |
^{27}C_{7} = ^{27}C_{r} |
A |
40 |
2 |
^{45}C_{5} = ^{45}C_{r} |
B |
2 |
3 |
^{8}C_{r} − ^{7}C_{2} = ^{7}C_{1} |
C |
20 |
જવાબ :
1-C, 2-A, 3-B
1 |
^{8}C_{r} − ^{7}C_{6} = ^{7}C_{5} |
A |
3 |
2 |
^{6}C_{r} − ^{5}C_{3} = ^{7}C_{2} |
B |
10 |
3 |
^{12}C_{r} − ^{11}C_{10} = ^{11}C_{9} |
C |
1 |
4 |
^{8}C_{r} − ^{7}C_{1} = ^{7}C_{0} |
D |
6 |
જવાબ :
1-D, 2-A, 3-B, 4-C
1 |
^{8}C_{2} |
A |
6 |
2 |
^{10}C_{3} |
B |
120 |
3 |
^{6}C_{5} |
C |
28 |
જવાબ :
1-C, 2-B, 3-A
1 |
C(6,2) |
A |
10 |
2 |
^{10}C_{2} |
B |
15 |
3 |
^{10}C_{1} |
C |
45 |
જવાબ :
1-B,2-C,3-A
-.
આ પ્રકરણને લગતા વિવિધ એનિમેશન વિડીયો, હેતુલક્ષી પ્રશ્નો, ટૂંકા પ્રશ્નો, લાંબા પ્રશ્નો, પરિક્ષામાં પુછાઈ ગયેલા પ્રશ્નો તેમજ પરિક્ષામાં પુછાઈ શકે તેવા અનેક મુદ્દાસર પ્રશ્નો જોવા અમારી વેબસાઈટ પર રજીસ્ટર થાઓ અથવા અમારી App ફ્રી માં ડાઉનલોડ કરો.
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.