Possible cases:

(i)BwinsA, B winsC=1/4

(ii)AwinsB,BwinsC,CwinsA,BwinsA,BwinsC=1/32

(iii)AwinsB,BwinsC,CwinsA,AwinsB,BwinsC,CwinsA,BwinsA,BwinsC=1/256

 

Clearly an infinite geometric series with first term=1/4;common ratio=1/8

Required probability=2/7

Clearly x+y=2n (where x>0,y>0)

Hence total no. of possible cases are C(2n-1,1)=2n-1....(i)

For P to lie on x=y there is only one possible casei.e. from for the required situation 2n-2/2n-1 is the required answer

 

The number of distinct quadratic equations of the type Ax²+Bx+C=0 that can be formed when A,B,C are selected from {1,2,3,4,5,6} is:

 

(i) 216   (ii) 198  (iii) 181  (iv)120

Let x⁴+ax³+bx²+cx+1=(x²+mx+1)(x²+nx+1)=0..........(i)

 

Clearly ,  a= m+n ; c= m+n ; b=2......(ii)

 

Since the mother equation above has only real roots so has her sons

 

Hence m²>=4   ; n²>=4 ........(iii)

 

Again a,b,c>0 i.e. the least positive values of a,b,c satifying the given conditions is

 

a=4 ;b=2;c=4   ..........[Since m=2,n=2 from (i),(ii)&(iii) ]

let there be 2n+1 stones i.e. n stones on each side of a single middle stones

since all stones r arranged the middle stone(which i m assuming not being done linearly along the same line), hence the man in total covers

n+ 2xn(n+1)/2+2x(n-1)n/2=n+n(n+1)+n(n-1)=n(1+2n)

i.e. 10xnx(2n+1)=3000

i.e. 2n^2+n-300=0