Combinatorics Problem - # of ways to form number How many ten-digit positive integers are there such that all of the
following conditions are satisfied:
(a) each of the digits 0, 1, ... , 9 appears exactly once;
(b) the first digit is odd;
(c) five even digits appear in five consecutive positions?
From Combinatorics by Pavle Mladenovic
My approach is as follows:
We first choose where to place the five even digits since that is the most restrictive condition. So they can be placed in slots 1-5 all the way to 6-10 (ex. 1st to 5th digits). There are 6 ways to do this, and then we have 120 ways (5!) to place the even digits in the slots, as well as 5! ways to place the odds digits, for a total of 6 * 120 * 120 ways. Wondering if this is accurate or if I'm miscounting something.
Hello everybody, was wondering if I can get some help solving this problem. Not sure how to approach it. Thank you in advance.
 A: Focussing on the 5 even numbers first is a good way to go. There are 5 different positions the set of them may go in (2,3,4,5 or 6 for the first even digit).
For each of the 5! * 5 arrangements of the even digits, there will be 5! different orders for the odd digits.
This is exactly the same as your method except you forgot the condition that the first digit must be odd. Therefore there is one less position for the consecutive even digits (5 not 6).
(You have 5!*5!*5 = 72,000 possible combinations)
I hope this helps.
A: The first digit must be odd, I use X to mark it. I use $[-----]$ to mark the five consecutive positions for even integers. I use $[~~]$ to mark other seat for odd integers. So we have $5$ ways to locate the five consecutive positions for even integers:
$$[X][-----][~~][~~][~~][~~]$$
$$[X][~~][-----][~~][~~][~~]$$
$$[X][~~][~~][-----][~~][~~]$$
$$[X][~~][~~][~~][-----][~~]$$
$$[X][~~][~~][~~][~~][-----]$$
Next, we have $5!$ ways to arrange the odd integers, and $5!$ ways to arrange even integers. The final answer is:
$$5\cdot 5!\cdot 5!=72000$$
