Let X_ n = 1,2,3, . n) and let a subset A of X_ n be chosen so that every pair of elements of A differ by at least 3. (For example, if n =5, ~A can be , 2 or 1,5 among others). When n=10 , let the probability that 1 A be p and let the probability that 2 A be q . Then

Step-by-step solution

When n =10 , Let A _ r be no. of ways of selecting r numbers No. of selection of A is = n ( A ₀ )+ n ( A ₁ )+ n ( A ₂ )+ n ( A ₃ )+ n ( A ₄ )=1+10+(7+6+5+ +1)+ (4+3+2+1)+(3+2+1)+(2+1)+1 +1=11+ 7.8 2 +10+6+3+1+1=60 N ( p )= n (no. of ways 1 is selected )=1+7+4+3+2+1+1=19 N ( q )= n (no. of ways 2 is selected )=1+6+3+2+1=13 So. p = 19 60 and q = 13 60 p - q = 1 10