How to determine the number of $4$-digit numbers that do not contain the sequence $17$

Question

I'm working on this:

Determine the number of 4-digit numbers that do not have the sequence 17. Eg 47113 an allowed 5-digit number, 1572 an allowed 4-digit number but 1723,3175,0254 are not allowed.

I'm stuck on how to find the probability of finding 17 in the number.

As suggested by @Nick, this problem can be conquered by a simple application of Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. — user2661923, Nov 30 '22 at 22:47

score 3 · Answer 1 · answered Nov 30 '22 at 22:38

3

There are 9000 4 digit numbers

To count the ones containing 17 take cases:

17XX - 100 cases
Y17X - 90 cases
YX17 - 90 cases
X is 0-9.  Y is 1-9

don't forget to subtract 1 for the double counted number 1717

answered Nov 30 '22 at 22:38

WW1

10,497

I sumbmitted this answer and it was not correct. Any suggestions why? Put in 8719. – Dec 01 '22 at 05:32
2

WW1 says 9000-100-90-90, so 8720. He says that we substract 1717 twice, when substracting 17xx, and xx17. So final count is 8720+1=8721. – Lourrran Dec 01 '22 at 12:38

How to determine the number of $4$-digit numbers that do not contain the sequence $17$

1 Answers1