5412 is the solution

please tell how?

I had the same question but with 28 objects instead . What would be the correct solution?. I got 2212

got 3989. when will the official solutions come.

got 32x113 (=3616) but for some reason i miscalculated it as 3614

for the same question as Mindjolt, I got 2268.

kgdpsdurg wrote: got 32x113 (=3616) but for some reason i miscalculated it as 3614 I think you are right......I ended up with the same result......

Yay atleast 1 person who got the same answer.

Method please?

total ways to select 3 out of 32 - no of ways in which 2 are adjacent - no of ways in which all 3 are adjacent - no of ways in which 2 are diametrically opposite + no of ways in which 2 are adjacent and 1 is diametrically opposite in short - principle of inclusion and exclusion

Can you please list out the calculations?

$\binom{32}{3} - 32*28 - 32 - 16*28 + 32$ it comes out to be $32*113 = 3616$ but mistakenly it wrote it as $3614$

Its $5772$ for $36$ objects in the Delhi/central paper.

anantmudgal09 wrote: Its 5772 for 36 objects in the Delhi/central paper. I got $5412$.....

For 36 objects.... What you got is for 32 ain't it?

Viswanath wrote: anantmudgal09 wrote: Its 5772 for 36 objects in the Delhi/central paper. I got $5412$..... I got $5412$ too.

kgdpsdurg wrote: Suppose 32 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite? This problem asks for 32 objects where 5412 is most likely correct. Mine was for 36 objects as in the WB/Delhi set.

For 2n objects the answer is 2nC3 - 2n(2n-4) -2n -n(2n-6) note that 2nC3= 2n(2n-1)(2n-2)/6

anantmudgal09 wrote: kgdpsdurg wrote: Suppose 32 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite? This problem asks for 32 objects where 5412 is most likely correct. Mine was for 36 objects as in the WB/Delhi set. I was talking about the $36$ one..........

Vishwanath is right. I am right. here is a solution for 32 Number of ways to choose 3 adjacent points : 32 Number of ways to choose 3 points such that only only two of them are adjacent : 32⋅28 Because there are 32 couples of adjacent points and the third point must not be adjacent the other two. Number of ways to choose 3 points such that two of them are diametrically opposite but not adjacent not adjacent- 16⋅26 Because the are 16 total diameters and you can choose the third point in 26 ways, as it must not be near the other two. Answer: (32C3)−32⋅28−32−16⋅26=3616

I believe this was the toughest question. Q5 was good but this was very tricky.

Okay, then I might have miscounted, let me see.

galav wrote: Vishwanath is right. I am right. here is a solution for 32 Number of ways to choose 3 adjacent points : 32 Number of ways to choose 3 points such that only only two of them are adjacent : 32⋅28 Because there are 32 couples of adjacent points and the third point must not be adjacent the other two. Number of ways to choose 3 points such that two of them are diametrically opposite but not adjacent not adjacent- 16⋅26 Because the are 16 total diameters and you can choose the third point in 26 ways, as it must not be near the other two. Answer: (32C3)−32⋅28−32−16⋅26=3616 exactly same solution.

Did the same as galav.....but did two major calculation blunders........ended up in 3800's. How much marks can i expect??

pjmaths wrote: Did the same as galav.....but did two major calculation blunders........ended up in 3800's. How much marks can i expect?? If u have committed only calculation blunders, then u can expect around 10 marks (idk if they even go thru your solutions properly, i have a very bad experience. u have that experience as well )

I have done 5 correct and 1 miscalculated 3810 ..do I consider myself in for inmo

laman560 wrote: I have done 5 correct and 1 miscalculated 3810 ..do I consider myself in for inmo Why not? you have a great chance!

Mine was also for 36 an elegant solution consists of the fact choose any object there are 36 ways now 4 choices are eleminated firstly there are two cases out of remaining 32 objects case 1 choose object next to an adjacent object to the first chosen naturally only 3 choices are eliminated in all other cases a further four will be eliminated similarly for the thir object then we need to divide by 6 as we are considering choices as different .I however miscalculated i think but i also got your answer

i have done 4 2 are 100% correct the infinte solution question in the delhi rmo I miscalculated in the p&c problem and have managed to prove the polynomial question

(32,3)-(32x30)+32 the number of 3 non adjacent points by principle of inclusion and exclusion is it right

now we substract the diametrical opposite pair (32,3)-(32x30)+32-(32x13)