Trouble Identifying Counting Problem Formula? Would this be the correct formula for the counting problem
Partition with Identical Items
$$
        \begin{pmatrix}
        n \\
        r  \\ 
        \end{pmatrix}=  \begin{pmatrix}
        n-1 \\
        r  \\ 
        \end{pmatrix} + \begin{pmatrix}
        n-1 \\
        r-1  \\ 
        \end{pmatrix}  \
$$
Moreover would anyone know a good example that might demonstrate a partition with Identical items? 
 A: I am a member of a group of $n$ people in a room, of whom $r \le n$ will be awarded
one of $r$ identical prizes. Number of ways: ${n \choose r}.$
But I am interested in whether I will get a prize. There are two
cases:  
I do not win a one of the prizes: Number of ways: ${n-1 \choose r}.$
Send me out of the room. Award the $r$ prizes among the $n-1$ people remaining.
I win one of the prizes: Number of ways: ${n-1 \choose r-1}.$
Start by giving me my prize. There are $n-1$ remaining people and $r-1$
remaining prizes.
Add the number of ways for the two cases to get the number of ways overall.
Note: This identity is the basis of 'Pascal's triangle'. (With no
disrespect to Pascal, there are reports of Indian, Arab, and Chinese manuscripts
showing this triangular configuration of numbers from many hundreds of
years before Pascal, some with no obvious indication of purpose.)    
A: For an example:
Suppose you had to choose a committee of 4 people, from a group of 7 people. Call these 7 people A, B, C, D, E, F, and G. Then you can choose this committee in ${7}\choose{4}$ ways. Another way you can calculate this is to split it up into two cases:
Case 1: A is part of the 4-person committee. There are ${6}\choose{3}$ ways to choose the other 3 people from the remaining 6 people to form the committee.
Case 2: A is not part of the 4-person committee. There are ${6}\choose{4}$ ways to choose the 4 people that make up the committee from the remaining 6 people.
We see that Case 1 and Case 2 account for all the 4-person committees once and only once. This is an example showing that:
$${{7}\choose{4}} = {{6}\choose{4}} + {{6}\choose{3}}$$
