# 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

## 2 Answers

for part a, it is $9 \times 10 \times 10$ as you did.

for part b, it is $9 \times 9 \times 8$

• so for part B I do not need to use the formal n!/(n-r)! ? To solve another part of the question which asked ... if repetition is not allowed? I did 10!/7! = 720. Is that correct? – Numb3ers Jan 31 '17 at 3:58

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.

Of course you can get to the same answer by just picking available digits in succession also, as you did, just needing to multiply together the digit-by-digit results for the final answer. You can perhaps see how the two processes are just the same in different formats.