@jain4444 said:getting 1/70why (4!/2!) two times ?
first arranging the 2 country and 2 coloured we want s0 4c2 *4!/2!
then araanging remaining 2 non flag poles and 2 remaining colured flags 4!/2!
2 times because either side is counted
A king has 100 sons. He takes every 10 sons to a jungle for hunting with no 10 being repeated. Let X be the no. of times that the king goes to the jungle and Y be the no. of times that each son goes to the jungle. Find X-Y
OA: 99C9 * 9
How many different values N can take if,
N = ± 8 ± 16 ± 24 ± 32 ± 40 ± 48 ± 56 ± 64 ± 72 ± 80
a) 56
b) 111
c) 881
d) 1024
N = ± 8 ± 16 ± 24 ± 32 ± 40 ± 48 ± 56 ± 64 ± 72 ± 80
a) 56
b) 111
c) 881
d) 1024
@soham2208 said:When x = 1/2 LHS = log(3) + 1 RHS = log(15) + log(2) = log(30) = log(3) + 1 LHS = RHS ..
my bad . calc.mistake ..
log5= 1- log2 .
log5= 1- log2 .
@sumit99 said:How many different values N can take if,N = ± 8 ± 16 ± 24 ± 32 ± 40 ± 48 ± 56 ± 64 ± 72 ± 80a) 56b) 111c) 881d) 1024
b.111
In a circle, 2 diameters AB and CD are drawn prependicular to each other
2 parallel chords PQ and RS are drawn parallel to diameter AB on either side of the circle such that distance of both the chords from diameter AB is r/2 where r is the radius of circle. The part of circle outside the chords containg points C and D [PCQ and RDS] are folded such that C and D meet at O
Find the are of the folded region
@rachit_28 said:A king has 100 sons. He takes every 10 sons to a jungle for hunting with no 10 being repeated. Let X be the no. of times that the king goes to the jungle and Y be the no. of times that each son goes to the jungle. Find X-YOA: 99C9 * 9
number of groups=100c10
number of groups with a unique son 99c9
x-y= 100c10-99c9= 99c9*9
number of groups with a unique son 99c9
x-y= 100c10-99c9= 99c9*9
@rachit_28 said:A king has 100 sons. He takes every 10 sons to a jungle for hunting with no 10 being repeated. Let X be the no. of times that the king goes to the jungle and Y be the no. of times that each son goes to the jungle. Find X-YOA: 99C9 * 9
X=100c10
Y=99c9
100c10-99c9
=99C9 * 9
Y=99c9
100c10-99c9
=99C9 * 9
@sumit99 said:How many different values N can take if,
N = ± 8 ± 16 ± 24 ± 32 ± 40 ± 48 ± 56 ± 64 ± 72 ± 80
a) 56
b) 111
c) 881
d) 1024
111 shayad
Total 10 terms
10*11/2 points on left of origin and 10*11/2 on right + origin
=111
@techsurge said:In a circle, 2 diameters AB and CD are drawn prependicular to each other2 parallel chords are drawn parallel to diameter AB of the circle such that distance of both the chords from diameter AB is r/2 where r is the radius of circle. The part of circle outside the chords containg points C and D are folded such that C and D meet at OFind the are of the folded region
pi*r^2/4
@rachit_28 said:Please explain the Y part.
consider symmetry .
every son would be in 1group for every 10group
--> 100c10/10 = 99c9
every son would be in 1group for every 10group
--> 100c10/10 = 99c9
@audiq7 said:@techsurge sry yar but didnt get the question. exactly kiska area nikalna hai?
edited for clarity
@pankaj1988 said:pi*r^2/4
nope edited question for better diagram
hope you will be able to solve better now