SOLUTION: How many nine-digit numbers can be formed using all the digits from 1 to 9, if the odd digits must always be in ascending order and the even digits in descending order?

Algebra ->  Permutations -> SOLUTION: How many nine-digit numbers can be formed using all the digits from 1 to 9, if the odd digits must always be in ascending order and the even digits in descending order?       Log On


   



Question 887963: How many nine-digit numbers can be formed using all the digits from 1 to 9, if the odd digits must always be in ascending order and the even digits in descending order?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Among the 9 digits 1,2,3,4,5,6,7,8,9, 
the 5 odd digits must appear 
in the order 1,3,5,7,9, left to right 
and the 4 even digits must appear 
in the order 8,6,4,2,left to right 

Of the 9 positions in the 9-digit number, we can choose
the 4 positions for the even digits in C(9,4) = 126 ways.

[Notice that although the even digits must be in a certain order,
order does not matter when we are picking the positions in which
the even digits will appear in the 9-digit number.  For example, in
the 9-digit number 123546789 we are picking positions 2,5,6, and 8 
for the even digits to go in.  Notice that picking positions 5,8,2, 
and 6 is the same set of positions as the set of positions 2,5,6, 
and 8 or positions 8,5,6, and 2. Therefore order of POSITIONS picked 
DOES NOT matter!  That's why we use combinations rather than 
permutations.  Hope this doesn't confuse you!]

The odd digits will be placed in ascendng order in the remaining 5 
positions in just C(5,5)=1 way.

Answer C(9,4) = 126 ways

Edwin