How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that…
How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?
12
18
36
72
Solution
The digits are 1,1,2,2,3,3,4,4. Odd digits {1,1,3,3} must go in the 4 odd positions; arrangements $= 4!/(2! \cdot 2!) = 6$. Even digits {2,2,4,4} must go in the 4 even positions; arrangements $= 4!/(2! \cdot 2!) = 6$. Total $= 6 \times 6 = 36$.