How many 3-digit numbers have exactly 2 identical digits?

[tatoo5ma]

Hey everyone
I had difficulties while solving this problem in probability.

We have 4 similar boxes, and 20 similar balls.
How many possible ways we can put those balls on those boxes, knowing that
-the first box contains 3 balls
- the second box contains 4 balls
- the third box contains 6 balls
- and the fourth box contains 7 balls

What I did, is since we have 4 boxes, it's going to be 4!
in the first box it's 3! and second, third and fourth i also 4!*6!*7!

So is the number of ways we can diperse those balls: 4!*4!*6!*7!
?

The secong problem stats:
How many 3-digits numbers we have that that contains 2 similar digits!
For this one, I have NO clue how to proceed.
Thanks for your help!

 

[BananaMan]

tried calculating the amount for 100->199 and then multiply?

remember that 110-119 all have 2 similar and then 101 121 etc but also each set of 10 has one such as 122 assuming that it is exactly 2 similar and not 2 or more

this is a very "ploddy" method but it will give you the correct answer, sure someone will come up with a better way soon

 

[mohlam12]

I think the number canbe written as:
100a+10b+10c

then you have 4 forms {xxy, xyx, yxx, xxx} (x =! y) and x is the number that repeats itself
than you may hav to analyse each form; for example, the first xxy
x has 9 possibilities (from 0 to 9)
the second x has one possibility, it should be esual to the previous x
and the y has 9 possibilities (from 0 to 10, minus that number which x took)
so it's, i think, 9*9*1 ?
and u'll do so for all the other forms

(if it says two similar, that means there may be 2 or 3 similar digits. however, if it said ONLY 2 similar, then it's 2 and nothin more)

i am sure if this is right, because we've not learned probability yet.

 

[tatoo5ma]

I guess I am able to solve the second problem...
But that first one, anyone lease can help? that would be appreciated :)

 

[hellraiser]

All the *balls* and *boxes* are similar. How are you supposed to differentiate between them. There will be only one arrangement possible to arrange the balls, I believe.