How to count the number of permutations?

Question

Mathematics

There are t, u and v (the number of ones, twos and threes). From them we compose a number of length (t + u + v digits).
How many different numbers can you get?
Do I understand correctly that this number will be equal to c( t, t+u+v ) * c( u, u+v )?
Where C(n,m) is the number of permutations n out of m

Answer 1

Look here - umk.portal.kemsu.ru/uch-mathematics/papers/posobie...
It seems to me that "Number of permutations with repetitions" is your case.
There is also an example below with chess pieces instead of numbers.

Answer 2

P = (t + k + v)!/(t! * k! * v!)
For example, take t = 1, k = 1, v = 2:
P = (1 + 1 + 2)!/(1! * 1! * 2!) = 4!/2 = 1 * 2 * 3 * 4/ 2 = 24/2 = 12 ways of permutations -- 12 different numbers can be obtained.
1) 1233
2) 2133
3) 3123
4) 3132
5) 2313
6) 2331
7) 3312
8) 3321
9) 1323
10) 1332
11) 3213
12) 3231