Find the number of all 6-digits numbers having exactly three odd and three even digits.
Problem
Source: RMO Mumbai 2016, P4
Tags: counting, combinatorics
TheOneYouWant
11.10.2016 10:54
The answer comes to be $281,250$. Solution: Case 1: First digit odd. This leads to $5$ choices for the first place, and now place the odd numbers in $10$ ways. Now, choices for the odd/even numbers are $5^5$. So, the total is $10.5^6$ Case 2: First digit even. Use the same logic as Case 1, to get $8.5^6$ Total: $18.5^6$
lifeismathematics
13.10.2022 09:10
so first of all notice if you choose any $3$ places surely that will occupy the unit digit of the number. now you can partition the places into group of two having $3$ spaces in which both groups go for odd or even. now there are $4\cdot 5^{5}$ ways to place one group having unit digit and numbers are even or there are $5\cdot 5^{5}$ ways for placing numbers such that group having unit digit places odd digits. now there $\binom{6}{3}$ ways to choose any three places out of $6$ but in this manner first place get arranged 2 times , so total ways to form the required number is $\frac{\binom{6}{3}(4\cdot 5^{5}+5\cdot 5^{5})}{2}=\boxed{18\cdot 5^{6}}$ ways