Math question: probability

[nkw5]

Can you help me understand this kind of problem?

If no digit appears more than once, how many 3-digit numbers can be formed from the digits 2, 3, 4, 5, 6, 8, 9?

I know the answer is 210. I can find the answer long hand (figure out each digit can make 30 3-digit numbers and then multiply by 7). I also know that 5x6x7 gets the answer; don't know why. Maybe that there are 7 numbers and if I use 3 numbers (same as the digits) counting backwards (7,6,5) and then multiply them I'll get the answer. I know this works for any similar problem, but I can't explain why.

 

[H2Andy]

is this a DIR admission test?

 

[redacted]

OK, How many choices do you have for the first digit? You have 7 ways to chose the first digit.

With that first digit gone, how many ways can you chose the 2nd digit? There are 6.

Now, with 2 digits gone, there are 5 digits left to chose the third digit.

Therefore, 7 x 6 x 5.

 

[Rick Murchison]

OK, How many choices do you have for the first digit? You have 7 ways to chose the first digit.

With that first digit gone, how many ways can you chose the 2nd digit? There are 6.

Now, with 2 digits gone, there are 5 digits left to chose the third digit.

Therefore, 7 x 6 x 5.

Beat me to it! Good job
Rick

 

[nkw5]

OK, How many choices do you have for the first digit? You have 7 ways to chose the first digit.

With that first digit gone, how many ways can you chose the 2nd digit? There are 6.

Now, with 2 digits gone, there are 5 digits left to chose the third digit.

Therefore, 7 x 6 x 5.

Thank you! I think I was almost there. Now I can sleep knowing I can explain it tomorrow to a 6th grader who is better at math than I am. (Don't tell her I said that; I'm suppose to be her teacher.)

H2Andy, you crack me up.

 

[Charlie99]

If you want to assign him some reading, the subject would be permutations, combinations, and factorial.

The problem you described is the permutation of 3 things out of 7.

7 factorial divided by (7-3) factorial.