Question

A four-digit number is to be formed with digits 0, 1, 2, 3, 4 and 5 (without repetition). What is the probability that this number will leave a remainder 1, being divided by 9?

Answer 1

Total number of numbers formed with digits 0, 1, 2, 3, 4 and 5 (without repetition) $=5\times 5\times 4\times 3=300$.
Numbers divisible by 9 $\Rightarrow$ Sum of digits = 9.
Among the digits, following combination will give sum of digits = 9 $\Rightarrow$0, 1, 3, 5 (since repetition not allowed and 4-digit numbers have to be formed); 0, 2, 3, 4.
With combination of 0, 1, 3 and 5 $\Rightarrow 3\times 3\times 2\times 1=12$.
With combination of 0, 2, 3 and 4 $\Rightarrow 3\times 3\times 2\times 1=12$.
Total number of numbers = 24.
Since there are 24 numbers which are divisible by 9 with the digits mentioned and without repetition, hence there will be 24 numbers which will leave remainder of 1 when divided by 9.
Hence Probability $=\frac{24}{300}=\frac{4}{50}$.