An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without replacement. The number of ways in which this can be done is ${\lambda }^2\left(7!\right)$ then $\lambda =$

Sum of 0,1,2,...,9 is 45, and 45 is divisible by 9 To form an 8 digit number divisible by 9 (without repetition) we can discard 0,9 or 1,8 or 2,7 or 3,6 or
4,5 (so that sum of remaining digits is divisible by 9) and form an 8 digit number by using remaining digits Required = $\mathrm{\angle }8+4\times (\mathrm{\angle }8-\mathrm{\angle }7)=36\times \mathrm{\angle }7={\lambda }^2\cdot \mathrm{\angle }7$ where $\lambda =6$