Correct Answer - B
`(b)` Five people can be divided into three groups in `(1,1,3)` or `(1,2,2)` ways
Hence, total number of ways
`=(5!)/(3!(1!)^(2))xx(1)/(2!)+(5!)/((2!)^(2)1!)xx(1)/(2!)`
`=10+15=25`
Correct Answer - A
The number of ways in which 20 people can be divided into two equal group is
`n(S) = (20!)/(10! 10! 2!)`
The number of ways in which...
Correct Answer - D
Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...
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)` Total no. of arrangement if all the girls do not seat side by side
`=["all arrangement"-"girls seat side by side"]`
`=8!-(6!xx3!)`
`=6!(56-6)=6!xx50`
`=720xx50=36000`
Correct Answer - A
`(a)` Required number of ways
`=("Number of ways in which" 16 "players can be divided in" `8` "couples")-("Number of ways when" S_(1) "and" S_(2) "are in the...
Correct Answer - C
`(c )` Let `C=` the event of the selected number being composite,
`E=` the event of there being no remainder.
`P(C )=(n(C ))/(n(S))=(15)/(25)=(3)/(5)`, `P(barC)=(2)/(5)`
`P(E//C)=(4)/(15)`, `P(E//barC)=(1)/(10)`
`P(E)=P(C...