# Homework Help: How many 5-digit briefcase combinations contain

1. Nov 17, 2013

### cronuscronus

Hi all. I wanted to double check some of my work, and get some feedback if there's any errors. Any help is appreciated!

a) A particular digit (say 5, for example).
Essentially we're choosing one digit. 10C1 is 10, and we have 5 slots, so the answer here is 10^5 = 100,000 combinations.

b) A pair and three other distinct digits. (e.g 27421)
Here we are really just choosing 4 distinct digits. So we calculate the combinations without replacement. 10*9*8*7 = 5,040 combinations.

c) Two pairs of distinct digits and 1 other distinct digit. (e.g
12215)
Here we are choosing 3 distinct digits. So we calculate the combinations without replacement. 10*9*8 = 720 combinations.

d) A palindrome of length 5 (e.g. 12321 , or 11111 or 12221)
The first digit can be selected from 1-9. The next digit can be selected from 0-9. The next digit can also be selected from 0-9. The last two digits are determined.

So we can calculate 9*10^2 = 900 5-digit palindromes.

Thanks for the help!

2. Nov 18, 2013

### Staff: Mentor

Does that make sense? All told, there are 100,000 different 5-digit numbers, with no requirements that any of them be any particular number.

3. Nov 18, 2013

### cronuscronus

Hi Mark. It makes sense in my head because if it can be any particular digit, then it can be any digit. Is this incorrect?

4. Nov 18, 2013

### Staff: Mentor

Yes, I believe that is incorrect. "Any particular digit" doesn't mean "any old digit at random." For example, if you're looking for all the numbers with, say, a 5 in the most significant digit, there are 104 different possiblilites.

5. Nov 18, 2013

### cronuscronus

Hi Mark. I think that makes sense. We essentially remove it from the number of digits since it was already chosen? The other 4 slots are just the combinations of the chosen digit?

6. Nov 18, 2013

### cronuscronus

How do the rest look to you? I think I'm getting it :).