3-digit even numbers that can be formed if one of the digits is 5 and the following digit must be 7

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Quant Warrior

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How many 3-digit even numbers can you form if one of the digits is 5 and the following digit must be 7 ?

I solved that question and I found out that there are 5 such numbers : 570, 572, 574, 576, and 578.

Solution in my book is like this:

Given that the number must have a 57 in it and should be even at the same time, the only numbers possible are 570, 572, 574, 576, and 578. Also, if there is no 5 in the number, you will get 360 more numbers.

I don't understand that line in bold. Please help me with this.
 
Quant Warrior said:
How many 3-digit even numbers can you form if one of the digits is 5 and the following digit must be 7 ?

I solved that question and I found out that there are 5 such numbers : 570, 572, 574, 576, and 578.

Solution in my book is like this:

Given that the number must have a 57 in it and should be even at the same time, the only numbers possible are 570, 572, 574, 576, and 578. Also, if there is no 5 in the number, you will get 360 more numbers.

I don't understand that line in bold. Please help me with this.
Click to expand...
I'm guessing you need to count 3-digit even numbers that contain 7 but not 5.
 
The way you wrote your solution seems fine.

To get the solution that the book refers to the problem has to be written as How many 3-digit even numbers can you form if WHEN one of the digits is 5 the following digit must be 7 ?

There are two cases to consider.

Case 1: There are no 5's.

Case 2: There is a 5

You found the number of cases for case 1. Now try for case 1.
 
I re-read your post. The solution the textbook gave agrees with your solution. They just go on to mention that if there were no 5 in the number (and this was allowed) then there would be 360 more numbers (I do not agree with 360). It is extremely strange that the book had that as their solution. Can you please show us a copy of the problem and the given solution as maybe you left out some details without knowing it. Thanks!
 
You didn't copy the problem correctly. Turns out the number does not have to have a 5 in it.
And the solution sounds like the author didn't understand the problem.
 
What you wrote as

How many 3-digit even numbers can you form if one of the digits is 5 and the following digit must be 7 ?​

should have been

How many 3-digit even numbers can you form SUCH THAT if one of the digits is 5, [THEN] the following digit must be 7 ?​

Exact wording is important!

The solution should have been something like this:

There are two cases. If the number contains a 5, then the number must have a 57 in it and should be even at the same time, so the only numbers possible are 570, 572, 574, 576, and 578. On the other hand, if there is no 5 in the number, there are 8*9*5 = 360 possible numbers, because the first digit can be anything but 0 and 5, the second can be anything but 5, and the third can be 0, 2, 4, 6, or 8. The total is 5 + 360 = 365.​
 

Dr.Peterson said:
What you wrote as

How many 3-digit even numbers can you form if one of the digits is 5 and the following digit must be 7 ?​

should have been

How many 3-digit even numbers can you form SUCH THAT if one of the digits is 5, [THEN] the following digit must be 7 ?​

Exact wording is important!

The solution should have been something like this:

There are two cases. If the number contains a 5, then the number must have a 57 in it and should be even at the same time, so the only numbers possible are 570, 572, 574, 576, and 578. On the other hand, if there is no 5 in the number, there are 8*9*5 = 360 possible numbers, because the first digit can be anything but 0 and 5, the second can be anything but 5, and the third can be 0, 2, 4, 6, or 8. The total is 5 + 360 = 365.​
Click to expand...




So second case means if there are no 5s and 7s in a number then this means that they won't come after each other, hence 360 such numbers are possible ?

Am I right ?
 
Quant Warrior said:
So second case means if there are no 5s and 7s in a number then this means that they won't come after each other, hence 360 such numbers are possible ?

Am I right ?
Click to expand...

If there are no 5's, then "come after each other" is meaningless; there might be a 7 anywhere, or none. The conclusion of the "if" can be ignored when the condition is false; there is no longer any condition on 7's.
 
I don't understand at all. I am able to understand the first case, but I am not able to grasp the second case at all.
 
Do you mean you don't understand the meaning of the description, or how to do this calculation to get the 360?

Please tell us whatever you do understand, so we can start there. Maybe point out which parts of post #8 you follow, and what confuses you about other parts.
 
The problem never said that there must be a 5 before 7. It said that IF there is a 5 then 7 must follow. It never said that you can't have 274 for example. If there is no 5 in the sequence then you have no restrictions (other than the first digit is not 0). So you can pick any 8 digits for the 1st digit, any of 9 for the 2nd digit and 5 for the 3rd digit. This gives 8*9*5 = 360 possible numbers that do not have 5 in it.

The case with no 5's is not hard. You just can't have a 5!
 
Jomo said:
The problem never said that there must be a 5 before 7. It said that IF there is a 5 then 7 must follow. It never said that you can't have 274 for example. If there is no 5 in the sequence then you have no restrictions (other than the first digit is not 0). So you can pick any 8 digits for the 1st digit, any of 9 for the 2nd digit and 5 for the 3rd digit. This gives 8*9*5 = 360 possible numbers that do not have 5 in it.

The case with no 5's is not hard. You just can't have a 5!
Click to expand...

Now I understood it perfectly. I was not getting the question in the first place. Your explanation of the problem and then an example made everything clear. Thank you so much. I was about to give up on this question, but you saved me.
 
Dr.Peterson said:
What you wrote as

How many 3-digit even numbers can you form if one of the digits is 5 and the following digit must be 7 ?​

should have been

How many 3-digit even numbers can you form SUCH THAT if one of the digits is 5, [THEN] the following digit must be 7 ?​

Exact wording is important!

The solution should have been something like this:

There are two cases. If the number contains a 5, then the number must have a 57 in it and should be even at the same time, so the only numbers possible are 570, 572, 574, 576, and 578. On the other hand, if there is no 5 in the number, there are 8*9*5 = 360 possible numbers, because the first digit can be anything but 0 and 5, the second can be anything but 5, and the third can be 0, 2, 4, 6, or 8. The total is 5 + 360 = 365.​
Click to expand...
Now I understand it very well after @Jomo explanation. Thanks a lot.
 
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