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AMIYA KUMAR

Maths By Amiya, PnC

11.11.11.11. Mom demanded you to purchase 10 fruits of exactly 3 different types. Then how

many different ways you can full fill her demand from a fruit market where you

found 12 different types of fruits enough to full fill demand of a village.

a. Infinite b. 12c3 c. 12c3*12c2 d. 12c3*9c2 e. NoT

Sol: a+b+c = 10 => 9c2*12c3

12.12.12.12. In how many ways we can arrange 20 people on a regular decagon such that people

are sitting on the vertex and middle of the sides

a. 19! b. 19!/2 c.2*19! d. None of these

Ans : 2*19!

13.13.13.13. There are how many times digit 5 come, if we write number from 189 to 5004

(including both)

a. 1460 b. 1462 c.1466 d. None of these

0000 to 4999

5 at unit place - 5*10*10 = 500

5 at 10th place - 5*10*10 = 500

5 at 100th place - 5*10*10 = 500

5 from 0000 to 4999 =

500+500+500 = 1500

000 - 199

5 at unit place - 2*10 = 20

5 at 10th place - 2*10 = 20

189 - 199

5comes only 1 place

5000-5004 = 5

Total number of 5 = [0000 to 4999]

- [000-199] + [189-199] +5 = 1500

- 40 +1 +5

= 1466

14.14.14.14. Total number of integral ordered pair solution of a*b*c=1001^2

a. 27 b. 27c3 c. 216 d.864 e. NoT

Sol 864

a*b*c=1001^2 = 7^2*11^2*13^2

Required solution = (4C2)^3(3c0+3c1)

Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya

AMIYA KUMAR

15.15.15.15. There are how many four digits numbers are in which all digits are prime and

when we add 5 in the number the resultant numbers also have all prime digits.

a. 64 b. 70 c. 76 d.80 e. NoT

Sol:

2,3,5,7

If last digit 2 , then total no = 4*4*4 = 64

If last digit is 7 and second last digit is 2 , = 4*4 = 16

Total =80

16.16.16.16. In how many different way one a put 12 different rings in his 5 fingers of one hand

such that each finger should have minimum 2 ring.

a. 12! b. 6P2*12! c. 6c4*12! d.12!/(4!)^3 e. NoT

Sol:[c]

a+b+c+d+e= 12

with minimum two condition, 6c2*12!

17.17.17.17. Total number of odd positive integral solution of a+b+c+d

Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya

AMIYA KUMAR

19.19.19.19. How many terms of ((13)^(0.5)*x - (3y)^(0.2)*y)^100 has integral

coefficient

a. 20 b. 10 c. 11 d. 100 e. None of these

Sol: [c]

2a + 5b = 100 , total 11 non negative integral solution.

20.20.20.20. If w+ 5x + 5y + 5z = N has 165 non negative integral solution then what is the

value of Max(N)

a. 50 b. 45 c. 49 d. 44 d. none of these

Note: 165 = 11c3 = 2c2 +3c2 +4c2 +..... +10 c2

Sol: 5*9 -1 = 44

Take

w+ 5x + 5y + 5z = 0

w+ 5(x + y + z) = 0 => w=0 , x + y +

z=0 no of solutions 2c2 distribution of x

+ y + z=0

w+ 5(x + y + z) = 5 => w=0 , x + y +

z=1 no of solutions 3c2 distribution of x

+ y + z=5 & => w=1 , x + y + z=0 no

of solutions 2c2 distribution of x + y +

z=0 ,

so total solution if N=5=5*1 is 2c2 +

3c2 .

and this solution will continue till 9

now take N=10 to 14

w+ 5(x + y + z) = 10

=> w=0 , x + y + z=2 no of solutions

4c2

=> w=5 , x + y + z=1 no of solutions

3c2

=> w=10 , x + y + z=2 no of solutions

2c2

total solution 2c2+3c2+4c2

check its till 4c2 then max value is 5*(4-

1) - 1=14

now take N=15 to 19

w+ 5(x + y + z) = 15

=> w=0 , x + y + z=3 no of solutions

5c2

=> w=5 , x + y + z=2 no of solutions

4c2

=> w=10 , x + y + z=1 no of solutions

3c2

=> w=15 , x + y + z=0 no of solutions

2c2

total solution 2c2+3c2+4c2+5c2

check its till 5c2 then max value is 5*(5-

1) - 1=19

question is till 10c2, then max is 5*(10-

1) - 1=44

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AMIYA KUMAR

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