Question

Compute the number of ten-digit numbers that contain only the digits 1,2,3 and where the digit 2 appears exactly twice.

I did the following but I don't know if its right, how should I approach this type of questions?
I thought that there are 9*9 ways you can move the 2s around and then in the 8 spots lefts you choose 1 from 1,3 so my answer would be 9*9*(2)^8 ?

Answer 1

The two 2's can "grab" 10C2 positions. That is going to leave you with the other 8 spots occupied with a varying # of 1's and 3's. Now, depending on whether or not you require at least one 1 and at least one 3, you can get 2 different answers. Let's say you can have no 1's and no 3's for now:

(10C2)(8C0 + 8C1 + 8C2 + 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8)

would be the answer in that case. If you have to have at least one 1 and at least one 3, then:

(10C2)(8C1 + 8C2 + 8C3 + 8C4 + 8C5 + 8C6 + 8C7)

Answer 2

The 8Cx terms come from "x" being the # of 1's (or 3's, take your pick). These remaining 8 spots are "grabbed" by the # of that digit.

Answer 3

ok I get the 10C2 but I still don't get the 8Cx terms and why do you add them ?

Answer 4

is it e.g 8C1 means there is only one 1 in the number so it only "grabs" one position and the rest are 3 , then plus 8C2 , meaning the combinations where there are two 1's two and the rest are 3s?

Answer 5

Yes, you are getting it. You came up with this as I was typing. What I was typing follows:

8C0 would mean that the remaining 8 spots have no 1's and there's only one way to have that: all 3's in those 8 remaining spots.

8C1 gives 8 places for that 1

8C2 for (8)(7)/(2) for 2 1's, etc. . . for up to 8C8

You add them because each situation "x number of 1's" is discrete. It's separate an distinct from all the other situations. It's like you are looking at each situation "in a bubble".

Answer 6

Making sense now @arykk ?

Answer 7

it makes sense now, thank you so much!

Answer 8

uw! Good luck to you in all of your studies and thx for the recognition! @arykk