@Faruq said:Two cases are possible a+b-ab=42 and ab-a-b =42Total 4 ways are -> (0,42)(2,-40),(44,2),(-42,0)
Only 1 of the two can be taken at a time! So, 2 pairs only
@ScareCrow28 said:( 25*21C1 + 25 )/25C3 = 11/46 ?
please explain the 2nd part..
25*21C1 is correct..i didn't get the 25 waala part in which I have to select 3 consecutive persons..
@VJ12 said:@ScareCrow28Areyy yes, but i used it and i got the same solution. Kuch exception hai kya?
Luck by chance 
Anyways, you can't use euler's here..ans matches or not!
Anyways, you can't use euler's here..ans matches or not!@heylady said:please explain the 2nd part..25*21C1 is correct..i didn't get the 25 waala part in which I have to select 3 consecutive persons..
You can do it by solving for 4 persons around a cirular table.. 1, 2, 3 and 4
Ways of selecting 3 = 123, 234, 341 and 412 = 4 ways
Correspondingly for 25 persons it will be = 25..( If i am not making a blunder )
What's the ans?
@Faruq said:In how many ways can 10 identical objects be distributed among 5 different boxes, such that exactly two of boxes have one each and the rest of the boxes could have any number of objects but not 1 object in them.
8 in 3
2,2,4->3
2,3,3->3
2,2,4->3
2,3,3->3
0,2,6->6
0,3,5->6
0,4,4->3
0,0,8->3
24*5c2=>240??
EDIT @Faruq
@Faruq said:How many pairs of integers exist such that the difference between their sum and product is 42?a) 1b) 2c) 3d) 4
(x + y) - xy = 42
x + y - xy = 42
x + y(1 - x) = 42
x + y(1 - x) - 1 = 41
(x - 1)(1 - y) = 41
x = 42...y = 0
x = 2...y = -40
2 pairs ?
@VJ12 said:@techgeek2050 Bhai, iska solution kar do na..a1+a2=...an/n>=(a1a2a3...an)^1/n but how did you get a1+2a2+...nan>=...
i guess question me it was a1 + 2a2 + 3a3 + .....nan
so (a1 + 2a2 + 3a3 + .....nan)/n >= (a1. 2a2. 3a3.....nan) ^(1/n)
(a1 + 2a2 + 3a3 + .....nan)/n >= (1.2.3....n)^(1/n) (a1.a2.a3.....an)^(1/n)
(a1 + 2a2 + 3a3 + .....nan)/n >= (n!)^(1/n) (k)^(1/n)
i hope that's clear :)
@Faruq said:In how many ways can 10 identical objects be distributed among 5 different boxes, such that exactly two of boxes have one each and the rest of the boxes could have any number of objects but not 1 object in them.
Boxes be a, b, c, d and e
a+b+c+d+e = 10
Two boxes 1 each = 5C2 ways = 10 ways
So, a + b + c = 8
___2__2__4 --> 3 ways
__2__3___3 --> 3 ways
__0__4___4 --> 3 ways
__0__2___6 --> 6 ways
__0__3___5 --> 6 ways
__0__0___8 --> 3 ways
Total = 10*(24) = 240 ? ..
Edited
@ScareCrow28 said:You can do it by solving for 4 persons around a cirular table.. 1, 2, 3 and 4Ways of selecting 3 = 123, 234, 341 and 412 = 4 waysCorrespondingly for 25 persons it will be = 25..( If i am not making a blunder )What's the ans?
sarkar isme btana zara kahan gadbad h
_a_b_c_
p+q+r+s=22
p q r have to be 1 atleast
p+q+r+s=19
22!/19!3!
21*20*19/6
25*4*23
1540
760/2300
76/230
38/115
p+q+r+s=22
p q r have to be 1 atleast
p+q+r+s=19
22!/19!3!
21*20*19/6
25*4*23
1540
760/2300
76/230
38/115
@ScareCrow28 said:You can do it by solving for 4 persons around a cirular table.. 1, 2, 3 and 4Ways of selecting 3 = 123, 234, 341 and 412 = 4 waysCorrespondingly for 25 persons it will be = 25..( If i am not making a blunder )What's the ans?
No OA...but the approach is correct...so blunder won't take place...
@mailtoankit said:(x + y) - xy = 42x + y - xy = 42x + y(1 - x) = 42x + y(1 - x) - 1 = 41(x - 1)(1 - y) = 41x = 42...y = 0x = 2...y = -402 pairs ?
what about x=2 , y= 44 ?
@Faruq said:Two-third of a consignment was sold at a profit of 5% and the remainder at a loss of 2%. If the total profit was Rs. 400, the value of the consignment (in rupees) was:a] 20000 b] 15000 c] 12000 d] 10000
15000 😉 ? This was surely the easiest asked by faruq 😉
@Subhashdec2 said:sarkar isme btana zara kahan gadbad h_a_b_c_p+q+r+s=22p q r have to be 1 atleastp+q+r+s=1922!/19!3!21*20*19/625*4*231540760/230076/23038/115
p/s can be 0 na?? One at a time..?..
@Faruq
@Faruq said:Two-third of a consignment was sold at a profit of 5% and the remainder at a loss of 2%. If the total profit was Rs. 400, the value of the consignment (in rupees) was: a] 20000 b] 15000 c] 12000 d] 10000
B
@ScareCrow28 said:Boxes be a, b, c, d and ea+b+c+d+e = 10Two boxes 1 each = 5C2 ways = 10 waysSo, a + b + c = 8___2__2__4 --> 3 ways__2__3___3 --> 3 ways__0__4___4 --> 3 ways__0__2___6 --> 6 ways__0__3___5 --> 6 waysTotal = 10*(21) = 210 ? ..Edited
bhai 0,0,8 wala case
Will edit again! :banghead:Find the last non-zero from right hand side of 120!