1. How many different positive, five-digit integers can be formed using the digits 2, 2, 2, 9 and 9?

2. I roll a fair 6-sided die five times. How many sequences of rolls will give me three 2's, one 4, and one 6?

whymenotsmart May 10, 2020

#1**+2 **

1 - There are 10 arrangements that can be made as follows:

{2, 2, 2, 9, 9}, {2, 2, 9, 2, 9}, {2, 2, 9, 9, 2}, {2, 9, 2, 2, 9}, {2, 9, 2, 9, 2}, {2, 9, 9, 2, 2}, {9, 2, 2, 2, 9}, {9, 2, 2, 9, 2}, {9, 2, 9, 2, 2}, {9, 9, 2, 2, 2} = 10 arrangements

Guest May 10, 2020

#3**+1 **

We could go ahead and count these directly, but instead we could count in general and then correct for overcounting. That is, if we had 5 distinct digits, there would be 5!=120 orderings. However, we must divide by 3! once for the repetition of the digit 2, and divide by 2! for the repetition of the digit 9 (note that if the repeated digits were different then we could rearrange them in that many ways). So, our answer is 5!/3! x 2!=10.

I understand now! This is the easier way, guest:)

whymenotsmart May 10, 2020

#4**+1 **

and i got #2, its 20 ways, since if there were no repeating digits, it would be 5!=120 ways, but there are three 2's, so 3! ways, so we just do 5!/3! to get 20 ways!!!:)

whymenotsmart May 10, 2020

#5**0 **

It is very true what you say. However, you should know that I have most these permutations and combinations programmed into my computer and it spits them out in microseconds and milliseconds. I, therefore, don't think about them at all as long as I enter the conditions that are required correctly.

Guest May 10, 2020