First let's get the correct answer.

There is one positive three digit integer that contains three 7's, 777.

Now for the ones that contain two 7's.

We could start with 77.

Digits: 7 7 9 Possible Ones Digits We already used 777. So 7 cannot be a ones digit.

We could split the 7's.

Digits: 7 9 Possible Tens Digits 7

We could end with 77.

Digits: 8 Possible Hundreds Digits 7 7

Total Three Digit Numbers That Include At Least Two 7's = 1 + 9 + 9 + 8 = 27

Now let's consider what you did.

Hmm. I see that you considered the numbers with one 7 among their digits, but aren't there other three digit numbers that don't have at least two 7's, melguy?

Sorry Marty I am really horrible with Counting and Probability questions. After a lot of work I could just grasp the basics and am really struggling with these 2 topics. I am thinking of maybe skipping anything that comes unless of course it's very basic!

melguy wrote:Sorry Marty I am really horrible with Counting and Probability questions. After a lot of work I could just grasp the basics and am really struggling with these 2 topics. I am thinking of maybe skipping anything that comes unless of course it's very basic!

melguy, what you missed doesn't have much to do with permutations. You did the permutations work that you did JUST RIGHT. However, you left out an entire category of integers.

With that in mind, figure out what you other category you have to subtract, and in so doing train yourself to get this type of question right, as you CLEARLY CAN.