I am trying to calculate the number of ways we can arrange the letters U,U,R. When I do this by hand the result is : 3 (UUP , UPU , PUU). But when I apply the formula which gives the number of ways to distribute k indistinguishable objects to n distinguishable boxes it doesn't work. The formula is C(n+k-1,k). To solve it using the formula I do n=3 because we have 3 spaces for the letters and k=2 because it only matters how we can arrange the Us. But the formula then is C(2+3-1,2)=C(4,2)=6 which is double the ways. Am I wrong somewhere?
PS. when I try the formula for 1 distinct object in 3 distinguishable boxes , I get the correct number 3. Why aren't the two things the same? 2 indistinguishable in 3 distinguishable the same as 1 distinguishable in 3 distinguishable
The formula for finding this is $$\frac{3!}{2!1!}=3$$or more generally $$\frac{n!}{k_1!\times\cdots\times k_m!}$$where $k_1+\cdots+k_m=n$.
This if there are $m$ distinct letters and letter labeled $i$ appears $k_i$ times and the arrangement counts $n$ letters in total.