3)18
EDIT: Sorry its 18. WE GET nC2 = 153. n= 18
Q1. Every one of the 10 available lamps can be switched on to illuminate a certain hall. The total number of ways in which the hall can be illuminated is -:
a) 55
b) 1023
c) 210
d) 10!
My Working : Each lamp can or cannot illuminate a hall, i.e., it has two states of working. So for n lamps, the hall can be illuminated in 210 ways. But the answer is 210 - 1....
Q2. If a_{n} = \sum_{r = 0}^{n}{\frac{1}{^{n}C_{r}}} then \sum_{r = 0}^{n}{\frac{r}{^{n}C_{r}}} equals?
Couldn't get this at all...
Q3. In a football championship, 153 matches were played. Every team played one match with each other. The number of teams participating in the championship is?
a) 17
b) 18
c) 9
d) none of these
Q4. The sides AB, BC, CA of a triangle ABC have 3, 4, 5 interior points respectively on them(interior and on them??). The total number of triangles which can be constructed using these points as vertices is :
a) 220
b) 204
c) 205
d) 195
Q5. All letters of the word EAMCET are arranged in all possible ways. The number of such arrangements in which no two vowels are adjacent to each other is :
a) 360
b) 144
c)72
d) 54
My Working : Total possible arrangements = 6!/2! as E repeats twice. Now vowels we have = E(2), A(1). So in 6 places, possible ways to arrange two vowels together = 5 x 2 = 10(EE and EA). So total ways = 6!/2! - 10 = 350 which is not the ans...
Thanks a lot terror...I was following the opposite way. This works just as well!
Ans 2) http://www.targetiit.com/iit-jee-forum/posts/binomial-ii-12752.html
bhai pritish the hall cannot be illuminated if all the lamps are off
so 1023
first.i tried like,taking the 3 vertices+interier points mentioned...but no option matched dude..
so i went for this .......is it correct or not?
I took separate cases, possibly that's where I went wrong.
(12C3 - 3C3) + (12C3 - 4C3) + (12C3 - 5C3)
That's all(in post #4)??? lol...my level of P&C is quite low.
Hey I used the same formula you used in post #5!
When there are n points in a plane and m are collinear, the number of triangles possible = nC3 - mC3
But won't there be separate cases? I got 3 x 12C3 instead of just 12C3, hence the wrong answer...
And why does the question ask for interior points which lie on the sides?? lol agar interior hain toh sides per kaise lie kar sakte hain...
q3.n=number of teams...
we need to just group 2 teams and see hw many combintions are possible..
nc2=153
n=18
Okay so 210 includes the case when all lamps are off..thanks. Baaki?