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 :).