Total number of numbers from given Condition =n(s)=26 Every required number is of the form A=7.(10a1+10a2+10a3+......)+111111 Here 111111 is always divisible by 21. ∴ If A is divisible by 21 then 10a1+10a2+10a3+...... must be divisible by 3. For this we have 6C0+6C3+6C6 cases are there ∴n(E)=6C0+6C3+6C6=22 ∴ Required probability =