Originally Posted by

**Vinod** Hello,

I calculated the answer for probability of two repetitions of random digits manually. A)If we take two digits, they can be written in 0001,0010,0100,1000. So total there are four ways. Now there are nine such pairs such as (0,1),(0,2)...(0,9) and ten random digits (0,1,2...,9).So there are total 4*9*10=360 numbers in which two repetitions of single digit can appear.

B)Let us take another case in which there are one repetition of one digit and again one repetition of another digit. e.g. take 0 and 1. we can write 0,1 in following 6 different ways. 0011,0101,1001,1010,1100,0110.There are such nine pairs(0,1),(0,2)...(0,9). Hence total such numbers are 9*6=54. For the nest pair 1 and 2,we have 48 numbers (6*8).So total we have 54+48+42+36+30+24+18+12+6=270.

Hence Total numbers of two repetitions of random digits are A)+ B)=630. Hence it's probability is $\frac{630}{10000}$.

Now, I cann't express my answer in binomial formulas. Would you express this answer in binomial formulas?