Combinatorial proof of $n! = {n\choose k}k!(n-k)!$

Question

Can someone give me some insight on the proof of $$n! = {n\choose k}k!(n-k)!$$ I understand algebraically why they are equal but I'm having trouble seeing what the right side is actually saying. On the right side I see that we're ordering $k$ objects multiplying it by the ordering of the number of objects that are left. I don't understand where the ${n\choose k}$ comes from. Any help is appreciated.

Answer 1

LHS is the total number of ways to order $n$ items in a row. The RHS counts the same, but as follows. First choose $k$ items of the $n$ without ordering them (that can be done in $n\choose k$ ways). Then, order the $k$ items you chose (that can be done in $k!$ ways), and finally, order the remaining $n-k$ items (that can be done in $(n-k)!$ ways). The product rule then says you multiply these numbers together.

Answer 2

LHS: You rearrange/permute $n$ different objects, so there are $n!$ ways.

RHS: You rearrange/permute $n$ different objects as follows. First, take $k$ object which can be done by ${n\choose k}$ ways and then you permute these $k$ objects which can be done by $k!$ ways. Finally, you also have $(n-k)!$ ways to rearrange the left objects. So, in total, there are ${n\choose k}k!(n-k)!$ ways.