# Want to know the method

Q1) In how many ways a triangle can be formed by choosing numbers from the set $\{1,2,3...,10\}$ .

Q2) I dont know the question exactly, But it says six ants are on the edge of a regular octahedron such that no two are on the same edge.They start moving on the adjacent edges, find the probability that no two ants arrive on the same vertex.

Q3) the figure is above, in how many ways can $A$ reach $B$?

Q4) Number of integral pairs $(x,y)$ satisfying $a+ib=(a+ib)^{2002}$ where $i=\sqrt{-1}$

Note by Tanishq Varshney
6 years, 1 month ago

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These All are questions from IPU CET 2015 which was held on 17th May,2015 @Tanishq Varshney , I too want to know the answers .For Q2 i think there's already a probem (similar) on b'ant, i just don't recall it correctly

- 6 years, 1 month ago

add me up on whatsapp with +2348029770297

- 6 years, 1 month ago

Who in the blue hell, r u jabroni , and i feel that no one is lunatic enough to add u or contact u 😝 becoz u r a complete stranger to me

- 6 years, 1 month ago

Q4) It is easy to see that $|a+ib|=1 or 0$. Therefore, the answer will consist of all the $2001^{th}$ roots of unity, as well as $0+0i$.

- 6 years, 1 month ago

Q3. This is a binary situation, either up or down path hence total ways are $\displaystyle {2} ^ {n - 1}.$

- 6 years, 1 month ago

Sir, Do you mind explaining? I can't understand the formula : $2^{n-1}$

- 6 years, 1 month ago

To reach B , the nth line must be compulsorily taken and hence the nth line has only one choice. But in the first (n-1) lines one can take the upper path or the lower path, hence each of these (n-1) lines has 2 possibilities. Thus by product rule , total ways $= 2^{n-1} \times 1 = 2^{n-1}$ ways. Try a similar problem here.

- 6 years, 1 month ago

- 6 years, 1 month ago

Try playing with small cases for motivation and ease. Then see if you can arrive at a general idea. I say, start with a set $A = \{1,2,3,4\}$.

- 6 years, 1 month ago

Dude, the triangles question. I think you can only do it by counting.

Take three cases: equilateral, isoceles and scalene.

Equilateral is simple.

Isoceles : Let the sides be $a$ and $b$ and let $a$ be the side which is equal. Then probably use Triangle Inequality.

Scalene Take two cases right triangles and normal triangles. Again triangle inequality must do the trick.

My method is very long. There must be a shorter way. Hope there is.

- 6 years, 1 month ago

- 6 years, 1 month ago

sorry guys i don't have their answers

- 6 years, 1 month ago