+0  
 
0
104
3
avatar+91 

How many nine-digit numbers can be made using each of the digits 1 through 9 exactly once with the digits alternating between odd and even?

 Jun 18, 2021
 #1
avatar+48 
+3

Final Answer: 

2880

 

 

Reasoning:

 

Odd numbers: 1,3,5,7,9

Even numbers: 2,4,6,8

 

So there are 5 odd numbers and 4 even numbers.

 

 

544332211

^ The number possibilities

 

For the first box, there are five possibilities, because there are five odd numbers.

For the second slot, there are four because there are four even numbers.

For the third, there are four possibilities, because there are four odd numbers left, since you can not use whatever was the first option.

For the fourth slot, there are three possibilities, because there are three even numbers remaining.

The pattern continues, until we are down to the ninth box, the unit digit, where there is only one possibility remaining.

 

Multiply what you get from the slots:

5*4*4*3*3*2*2*1*1 

= 2880

 

Please reply if you find a mistake lurking!

 

=^-^=

 Jun 18, 2021
edited by TheOddOne  Jun 18, 2021
 #2
avatar+114485 
+3

o e o e o e o e o  

For it to alternate it must start and finnish with an odd    

5 odd places and 4 even places

5!*4! = 2880 

Just as TheOddOne already told you.   laugh

 Jun 18, 2021
 #3
avatar+91 
+1

i got this.

There are five odd digits and four even digits to be used. Because the digits must alternate between odd and even, this means that there is only one possible way to distribute odds (O) and evens (E): OEOEOEOEO. Now, there are  5! =120 ways to arrange the odd numbers, as there are five choices for the first slot, four for the second, and so on. Similarly, there are  4!=24 ways to arrange the even numbers. Our final answer is the product of  120 and 24, which is 2880.

And i find this logic clearer, Why?

 Jun 22, 2021

43 Online Users

avatar
avatar