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 $i=\sqrt{-1}$

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestQ3. This is a binary situation, either up or down path hence total ways are $\displaystyle {2} ^ {n - 1}.$

Log in to reply

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

Log in to reply

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.

Log in to reply

@Raghav Vaidyanathan @Andrei Golovanov @Nihar Mahajan @Vishwak Srinivasan

Log in to reply

sorry guys i don't have their answers

Log in to reply

@Ronak Agarwal @Mvs Saketh @Nishant Rai @Brian Charlesworth sir

Log in to reply

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

Take three cases:

equilateral,isocelesandscalene.Equilateralis simple.Isoceles: Let the sides be $a$ and $b$ and let $a$ be the side which is equal. Then probably use Triangle Inequality.ScaleneTake 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.

Log in to reply

Try playing with small cases for

motivationand ease. Then see if you can arrive at a general idea. I say, start with a set $A = \{1,2,3,4\}$.Log in to reply

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 $0+0i$.

Log in to reply

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

Log in to reply

add me up on whatsapp with +2348029770297

Log in to reply

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

Log in to reply