There are 10 stations on a circular path. A train has to stop at 3 stations such that no two stations are adjacent. The number of such selections must be: (A) `50` (B) `84` (C) `126` (D) `70`
A. `50`
B. `60`
C. `70`
D. `80`
Correct Answer - A
`(a)` Total selections `="^(10)C_(3)=120`
Number of selections in which `3` stations are adjacent `=10`
Number of selections in which `2` stations are adjacent `=6`
But there are `10` such pairs.
`implies` Total invalid selections `=10+6xx10=70`
Correct Answer - D
`vecV_(TG)=54xx(5)/(18)=15m//s=15hatj`
`vecV_(MT)=-18xx(5)/(18)=-5m//s=-5hatj`
`vecV_(MT)=vecV_(MG-vecV_(TG)`
`vecV_(MG)=vecV_(MT)+vecV_(TG)=-5hatj+15hatj`
`=10hatj=10m//s` along north
Correct Answer - D
`(d)` `a_(2)=(a_(1)+a_(2))/(2)`
`a_(3)=(a_(2)+a_(4))/(2)`
`a_(1)=(a_(2)+a_(2012))/(2)`
`a_(2012)=(a_(2001)+a_(1))/(2)`
Now `a_(2)+a_(4)+…+a_(2012)=3018`…….`(i)`
`2a_(2)+2a_(4)+..+2a_(2012)=6036`
`:.a_(1)+a_(2)+a_(3)+a_(5)+...+a_(2011)+a_(1)=6036`
`:.2(a_(1)+a_(3)+...+a_(2011))=6036`
`:.a_(1)+a_(3)+...+a_(2011)=3018`.........`(ii)`
By adding `(i)` and `(ii)` we get
`a_(1)+a_(2)+a_(3)+...+a_(2012)=6036`
Correct Answer - A
`(a)` The other five letters (other than `G`, `R`, `L`, `A`, `S`, `N`) in `11` places can be arranged in `"^(11)P_(5)` ways.
Then three remains six places....
Correct Answer - A
`(a)` First arrange `12` persons `A_(4),A_(5),….A_(15)` in `"^(15)P_(12)` ways
There remains `3` places. Keep `A_(1)` in the first place and arrange `A_(2)`, `A_(3)` in the remaining two...
The diagonal of the parallelogram made by two vectors as adjacent sides not passing through common point of two vectors represents triangle law of vector addition.