The number of ways of distributing `3` identical physics books and `3` identical methematics books among three students such that each student gets at least one books is
A. `45`
B. `55`
C. `64`
D. `72`
Correct Answer - C
The required probability is
1- probability of getting equal number of heads and tails
`=1-""^(2n)C_(n)((1)/(2))^(n)((1)/(2))^(2n-n)`
`=1-((2n)!)/((n!)^(2))xx(1)/(4^(n))`
Correct Answer - B
Let `P(m),P(c),P(c)` be the probabilaty of selecting a book of maths, physics, and chemistry, respectively. Clearly,
`P(m)=P(P)=P(c)=1/3`
Again let `P(s_(1))and P(s_(2))` be the probability that he solves...
Correct Answer - C
`(c )` `B_(1)B_(2)B_(3)…B_(7)B_(8)B_(9)B_(10)`
`(i)` When two terminal books are taken (`B_(1)B_(2)` or `B_(9)B_(10)`) then number of ways `=2xx7=14`
`(ii)` When two consecutive terminal books are not taken...
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...
Correct Answer - B
`(b)` Division of objects can be `(0,1,4)` or `(0,2,3)`
So total number of distribution ways
`=(5!)/(0!*1!*4!)xx3!+(5!)/(0!*2!*3!)xx3!`
`=30+60=90` ways
Correct Answer - C
`(c )` `m(S)=(12!)/(4!4!4!3!)3!=(12!)/(4!4!4!)`
`n(E)=` the number of ways in which any one boy gets both the pens
`="^(3)C_(1)^(10)C_(2)(8!)/(4!4!2!)2!=3xx45(8!)/(4!4!)`
`:.P(E)=(3xx45(8!)/(4!4!))/((12!)/(4!4!4!))=(2)/(11)`
`:.` Required probability `=1-(2)/(11)=(9)/(11)`
