Let A=x1,x2,x3,x4,x5 and f:A→A . The number of bijective functions such that fxi=xi for exactly two of the xi′s is (i=1 to 5)

Number of ways of two elements follow fxi=xi is 5C2 Rest 3 unit does not satisfy fxi=xi= Number of de-arrangement of 3 things =3!1−11!+12!−13!=3!12−16=6×3−16=2 Required no of ways =10×2=20