A simple combinatorics problem

[Yashbhatt]

Homework Statement

How many 4-digit odd numbers are there? (Without repetition)

Homework Equations

The Attempt at a Solution

I thought in the following way: For the last digit, we have 5 possibilities(1,3,5,7,9). For the last second we are left with 9 possibilities, 8 for the next one and 6 for the next one( because 0 is not allowed). So, I got 9 x 5 x 8 x 6 = 2160. But the answer is 2240.

Another approach: 5 for the 4th one, 8 for the 1st, 8 for the 2nd and 7 for the 3rd which gives the correct answer.

But I don't understand why the first method gave the wrong answer. I guess it is something like when the first 3 digits are odd, the last one has only 2 ways of filling, but I am not sure.

 

[PeroK]

Homework Statement

How many 4-digit odd numbers are there? (Without repetition)

Homework Equations

The Attempt at a Solution

I thought in the following way: For the last digit, we have 5 possibilities(1,3,5,7,9). For the last second we are left with 9 possibilities, 8 for the next one and 6 for the next one( because 0 is not allowed). So, I got 9 x 5 x 8 x 6 = 2160. But the answer is 2240.

Another approach: 5 for the 4th one, 8 for the 1st, 8 for the 2nd and 7 for the 3rd which gives the correct answer.

But I don't understand why the first method gave the wrong answer. I guess it is something like when the first 3 digits are odd, the last one has only 2 ways of filling, but I am not sure.

In your first attempt you forgot that if 0 was one of the previous digits, then there are 7 possibilities for the first digit. It's only 6 if you haven't yet had a 0.

 

[Orodruin]

It is not certain that you will have only 6 possibilities for the first digit in your original approach. If 0 was used in one of the middle positions, then you would have 7 possibilities.

 

[Yashbhatt]

So, that means I have to necessarily do it with the second method, right?

 

[PeroK]

So, that means I have to necessarily do it with the second method, right?

You don't have to do it that way, although it's almost certainly the simplest.

To rescue your original approach, you'd have to consider 2 cases: where 0 was one of the middle digits; and, where 0 was not one of the middle digits. You'd get the right answer, but it would be more complicated.

 

[Yashbhatt]

Okay. Now I get it.

But also won't we have to consider different cases when the first three are odd, and two are odd etc.? Because that changes the no. of ways in which the last digit can arrange itself.

 

[Orodruin]

Okay. Now I get it.

But also won't we have to consider different cases when the first three are odd, and two are odd etc.? Because that changes the no. of ways in which the last digit can arrange itself.

Yes, this is why it is most convenient to start with the last digit and then the first (the first cannot be zero or the one chosen for the last digit, which is not zero since it is odd).

 

[Yashbhatt]

Okay.