# Calculating the probability of a 6 digit phone number with no repeats

I have a stats problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track. Any assistance would be highly appreciated.

Q: What is the probability that a six-digit telephone number has no repeated digits? Do not allow the number to start with a zero.

So I know that there are 10 possible digits that can be selected, each time the pool of selected digits gets smaller.

The first digit cannot be zero therefore there are 9 possibilities for this digit. Then, seeing as the order doesn't matter, but repeats do, so I thought a permutation would be the correct method to apply.

9 * ((9!)/(9-5!)) = 136,080 (<-- total number of 6 digit numbers)

Without repeats and no 0 as first digit

Total number of 6 digit numbers = 10^6

Therefore the probability of this scenario happening is:

(139,080/10000000) = 0.13608

Did I make a mistake when using the permutation?

I guess that is a mistake in the solution.

If you calculate: (10*9*8*7*6*5)/(10^6) = 0.1512

My guess is, that the person, who did the solution forgot, that there are no leading zeros allowed.

The probability of a 6 digits number with no repeats is basically the number of permutations of 6 digits with no repetitions among the set of all possible numbers, so the probability is given by $P(E) = \dfrac{P(10,6)}{10^{6}} = 0.1512$.

The original answer is actually correct. The total number of 6 digit numbers is not 10^6, but 9*10^5 (as 011111 is actually a 5 digit number, so zeros should be excluded from the denominator as well).

9*9*8*7*6*5/(9*10^5) = 0.1512