# How many three digit sequences using the numbers 0 … 9 with conditions

How many three digit sequences using the numbers 0 ... 9 with conditions
(a) Repetition of digits is allowed and the sequence cannot start with 0?
(b) Repetition of digits is not allowed and the sequence cannot start with 0?

For part (a) I did the following:
(9)(10)(10) = 900

I am stuck on part (b), I know that when you sample without replacement you want to use n!/(n-r)! But in this problem n isn't constant. The first time a digit is picked there are 9 possible digits because the sequence cannot start with 0. The second time a digit is picked there are still 9 possible digits because we are sampling without replacement and the third time a digit is picked there are 8 possible digits left. I am not sure how to combine this information to solve for part (b). Any suggestions?

• Take the product like you did in part a. – Lanier Freeman Jan 31 '17 at 3:49

for part a, it is $9 \times 10 \times 10$ as you did.
for part b, it is $9 \times 9 \times 8$
For part (b), using $\dfrac{n!}{(n-r)!}$ will get you $\dfrac{10!}{(10-3)!}=\dfrac{10!}{7!} = 10\cdot 9\cdot 8$ options, ignoring the leading zero restriction. Then $\frac{1}{10}$ of these options will have a leading zero, so we can multiply by $\frac {9}{10}$ to get back to the valid solutions, $\frac {9}{10}\cdot 10\cdot 9\cdot 8 = 9\cdot 9\cdot 8 = 649$ options.