# Permutation and combination

1. Oct 8, 2012

### xiphoid

1. The problem statement, all variables and given/known data
how many 4 digit numbers are there which do not contain more than 2 different numbers?

2. Relevant equations

3. The attempt at a solution
all can contain the same digit
any 3 out of the 4 places can be occupied by the same digit
and
any 2 out of the 4 places can be occupied by the same digit

2. Oct 8, 2012

### leej72

You are on the right track, think about it this way. You have a choice of choosing the numbers from 1 to 9 and then you have three cases:

1. the case where all 4 numbers are the same number as you chose

2. the case where 3 numbers are the same number as you chose as well as you have to choose another number that's different from the other three

3. the case where 2 numbers are the same number as you chose and you have to choose 2 other numbers where they are different from each other and different from the other 2 you previously chose

Does that make sense?

3. Oct 8, 2012

### rcgldr

Shouldn't that be digits from 0 to 9 (a total of 10 possible digits)?

4. Oct 8, 2012

### Ray Vickson

Probably yes, but (presumably) the first (left-most) digit cannot be 0. (Since this is not made clear, if I were doing the problem I would solve both versions.)

RGV

5. Oct 8, 2012

### rcgldr

As you mention, it's not clear if leading zeroes are allowed, like a combination lock with 4 digits. I would assume they are allowed, since otherwise it eliminates all the combinations such as 0xxx, 00xx, 000x, 0000.

6. Oct 8, 2012

### leej72

Yes that was a typo, I did mean from 0 to 9 but I am not too sure if the first number would be allowed to be a 0. I would personally think otherwise, not letting the first number to 0

7. Oct 8, 2012

### xiphoid

as per the answer given, you will need to consider both the cases!