@[562114:Budokai001]
Yes you are correct.
11985 is what u must be getting.
Can u pls elaborate on the concept.
TIA
@mayankdangi said:@Budokai001Yes you are correct.11985 is what u must be getting.Can u pls elaborate on the concept.TIA
Buddy ,there are 18 non collinear points and 7 collinear points
First i find all possible quadrilaterals that is 25C4
But there are 7 collinear points
How can 4 points on a same line form a quadrilateral ?
Also How can 3 points on a same line form a quadrilateral? It is a triangle .
So we must subtract these 2 cases
case 1:4 points on a same line =7C4
case 2 : 3 points on a same line& other point on a non collinear point
7C3*18C1
so subtract these two cases to get the answer
@[562114:Budokai001]
Thanks, didnt think of case 2.
@ayushbhalotia said: Hi Puys!!Can some1 pls. help with the solution for the below mentioned questions:-Q1) Find the sum of all possible whole number divisors of 720?A) 2012 B) 2624 C) 2210 D) 2418 E)2520Q2) Find the sum of the series:- 1/1*5 + 1/5*9 + 1/9*13 +......... +1/221*225 A) 28/221 B)56/221 C)56/225 D) None of theseQ3) What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 .... + (1152!)^3 is divided by 1152? A) 125 B)225 C)325 D)205thanks in advance !!! Regards, Ayush
@[361226:ayushbhalotia]
2.) C. 56/225 (multiply by 4 and divide by 4 .. all terms except first and last will cancel out)
3.) B. 225 (4! ^3 is divisible by 1152 so take all terms less than that)
1.) solving.............
@ayushbhalotia said: Hi Puys!!Can some1 pls. help with the solution for the below mentioned questions:-Q1) Find the sum of all possible whole number divisors of 720?A) 2012 B) 2624 C) 2210 D) 2418 E)2520Q2) Find the sum of the series:- 1/1*5 + 1/5*9 + 1/9*13 +......... +1/221*225 A) 28/221 B)56/221 C)56/225 D) None of theseQ3) What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 .... + (1152!)^3 is divided by 1152? A) 125 B)225 C)325 D)205thanks in advance !!! Regards, Ayush
for 1. ) D 2418 ??
@[361226:ayushbhalotia]
@ @[475306:pulkitgurditta]- - for Q1) even i had got 2418 but the ans is B)2624.
For Q2 how did u get the logic that it has to be dividedand multiplied by 4??? is it because the diff between the numbers is 4?? pls reply..
@ayushbhalotia said: @ @pulkitgurditta- - for Q1) even i had got 2418 but the ans is B)2624.For Q2 how did u get the logic that it has to be dividedand multiplied by 4??? is it because the diff between the numbers is 4?? pls reply..
@[361226:ayushbhalotia] dude.. pehle kahin pe dekha tha maine ye...to solve kar diya
@[75523:abhijit111] Can anyone solve this question for me??
The number of circles that can be drawn out of 10 points of which 7 are colliner??
@[591492:Rithu] (7C1*3C2) + (7C2*3C1) + 1 = 85
right?
@[402270:krishna581] Answer is right!! but why should i have 3 points in order to construct a circle?? why cant i do it with two??
@[402270:krishna581] Can you explain it for me?? 
@[591492:Rithu] Theoretically infinite number of circles can be drawn that pass through two points. But if we take three non-collinear points, only one cirlce can be drawn. Hope you got it!
@[402270:krishna581] Got it...!! Given three points, it is always possible to draw a circle that passes through all three(this is d explanation rite ?? 
) So another way is 10C3-7C3=85 
@[591492:Rithu] Three points should be non-colilnear dude! otherwise not.
@[402270:krishna581] Dat's Obvious 
@[591492:Rithu] any other questions??
@[402270:krishna581] In how many ways a cricketer can score 200 runs with fours and sixes only?
@[402270:krishna581] Please do explain..
@[591492:Rithu] 17 ways?
@[402270:krishna581] Yup!! But howwwwwwwwww????????