bhai log binomial sikhne ki koi achchi link hai kya.....
@gupanki2 said:In a race A beats by 10 sec, B beats C by 20 seconds. By how many seconds did A beats c?
something is missing in question, otherwise ans should be more than 30
@gupanki2 said:@rahulgaur666 the question is correct bosss..... and i wonder shouldnt the answer be less than 30?
nope dude bcoz A's speed > speed of B > speed of C
@Asfakul said:In How many ways 8 distinct balls can be divided among 3 people such that anyone can get any number of balls? OA- 3^8Why cant we not perceive the problem as A+B+C=8where A,B,C are the number of balls each may receive. In that case number of solution is 8+3-1C2=10C2Please explain why cant we take this approach and where can we use this approach as well .@chillfactor@Subhashdec2@grkkrg@gs4890 PLEASE HELP !
because the balls are distinct.
@rahulgaur666 said:how come 16
x^2- y^2= + and - 247 solving this you get the following 16 sets: (16,3), (16,-3), (124,123), (124,-123), (3,16), (-3,16), (123,124), (-123,124), (-3,-22), (-3,16) and so on till 16 sets.
solve it like this:
factors of 247 are 1, 13, 19
take each value as either the sum of x&y; or difference of x,y. you will get the solution.
there is another way by grapgh which is comparatively easy.
i hope this helps.
@rahulgaur666 dude thats why it shoud be less than 30
when 'A" completed the race , gap b/w B and C should be less than 20
@zuloo said:x^2- y^2= + and - 247 solving this you get the following 16 sets: (16,3), (16,-3), (124,123), (124,-123), (3,16), (-3,16), (123,124), (-123,124), (-3,-22), (-3,16) and so on till 16 sets.solve it like this:factors of 247 are 1, 13, 19take each value as either the sum of x&y; or difference of x,y. you will get the solution.there is another way by grapgh which is comparatively easy.i hope this helps.
I thought same i.e. taking + 247 or -247 but then didn't take -ve value
@rahulgaur666 said:I thought same i.e. taking + 247 or -247 but then didn't take -ve value
it's a mod function even if you take -ve will work as the value becomes +ve.
@gupanki2 said:@rahulgaur666 dude thats why it shoud be less than 30when 'A" completed the race , gap b/w B and C should be less than 20
Hmm may be u r correct , mine mind isn't working , 
@Logrhythm said:urmm....let me explain with a smaller numberHow many of the first 15 numbers are coprime to either 3 or 5the thing here to notice is Either prime to 3 or 5Let's do it manually first:Coprime to 3: 10 numbersCoprime to 5: 12 numbersCoprime to 15 (both): 8 numbersso number coprime to either 3 or 5 are 5,10,3,6,9 and 12 => 6 numbersor 15*(2/3) + 15*(4/5) - 2*15*(2/3)*(4/5) = 10+12-16 = 6 Why does this occur is simply because of Either Or...and when we apply set theory, we are double counting it...so that is why we need to subtract it twiceHope it is clear
bhai what if there are 3 numbers, like in this questions
how many of the first 2100 natural numbers are either prime to 6,15 or 35 ??
This is what i did..
After adding Coprime to 6, Coprime to 15, Coprime to 35 , we have to take them in pairs. The factors are 2,3,5 and 7. so we'll have to subtract twice the coprimes of 30, 105 and 42 and thrice the coprimes of 210, am i right? not sure though..
@gupanki2 said:In a race A beats by 10 sec, B beats C by 20 seconds. By how many seconds did A beats c?
20/3 secs imo.
How many words can be made with word "SOFTWARE" in which vowels won't be together ?
@jain4444 said:How many words can be made with word "SOFTWARE" in which vowels won't be together ?
total number of arrangements = 8!
number of ways in which all 3 vowels are together = 6! 3!
number of ways in which any two vowels are together = 7! 2! 3
number of ways in which vowels are not together = 8! - 6! 3! - 7! 2! 3 = 5760 ??
@jain4444 said:How many words can be made with word "SOFTWARE" in which vowels won't be together ?
is it 2400 ??
@techgeek2050 said:bhai what if there are 3 numbers, like in this questionshow many of the first 2100 natural numbers are either prime to 6,15 or 35 ??This is what i did.. After adding Coprime to 6, Coprime to 15, Coprime to 35 , we have to take them in pairs. The factors are 2,3,5 and 7. so we'll have to subtract twice the coprimes of 30, 105 and 42 and thrice the coprimes of 210, am i right? not sure though..
bhai i guess the answer here wld be 700
2100*2/3*1/2 + 2100*4/5*2/3 + 2100*6/7*4/5 - 2((2100*2/3*4/5*1/2 - 2100*6/7*2/3*4/5*1/2) + (2100*4/5*6/7*2/3 - 2100*6/7*2/3*4/5*1/2)) - 3*2100*6/7*2/3*4/5*1/2 = 700
but why do questions like these? this, i am sure won;t feature in any of the MBA entrance exams, IIT mein bhi nahi aayega ;)