I'm stupid can someone help me understand this, please?

[James.LaCroix]

The digits of the number 522789 are arranged so that the resulting number is odd. Find the number of ways in which it can be done?

 

[James.LaCroix]

The digits of the number 522789 are arranged so that the resulting number is odd. Find the number of ways in which it can be done?

Never mind, I understood it.

Assuming the last digit(Red) cannot move and only the first five numbers(Green) can move. 522789
The number of combinations we can generate will be 5! or 5 x 4 x 3 x 2 x 1 which gives us 60 combinations if the numbers were unique.

But since not all the numbers are unique 522789 (the blue numbers are unique and the orange ones are not in this example), we need to make sure to remove the duplicates from the combinations, we can do this by simply dividing the 60 combinations by the number duplicates. Here we have two 2's. This means we have two duplicates. So, we will need to divide the 60 combinations by 2, which will give us 30.
Let's suppose we had these digits to work with, 552229. The number of duplicates would've been 2 (because there's two 5's) and 3 (because there's three 2's)

And finally, knowing that we have 3 Odd numbers (5, 7, 9) by which the combinations could end, we can simply multiply the number of combinations by 3 which gives 180 different combinations.

Answer: (5! / 2!) x 3 = 180

 

[Steven G]

Looks good to me.

 

[HallsofIvy]

And, please, do not ever say "I am stupid". You can, of course, think it. I do that all the time!

 

[JeffM]

Here is another way.

How many ways can we select the final digit so that the number is odd? 3.

How many pairs of positions can the 2’s fill?

[MATH]\dbinom{5}{2} = 10.[/MATH]
How many ways can we fill the remaining three positions? 3!?

[MATH]3 * 10 * 3 * 2 = 180.[/MATH]