We know that the multiplication principle states that if there are $a$ ways of doing $A$ and $b$ ways of doing $B$ once $A$ has happened, then there are $ab$ ways of performing both actions.

Keeping this in mind, suppose that I have a group of $n$ distinct objects and I want to choose $a$ of them. This can obviously be done in
$$\binom{n}{a}$$ ways.

However, it would seem that I could also first choose $a-b$ objects followed by another $b$ objects from the group of $n$ objects. Now, the number of ways of choosing the first $a-b$ will clearly be
$$\binom{n}{a-b}$$ and once they've been chosen the number of ways of choosing the remaining $b$ items will be $$\binom{n-(a-b)}{b}.$$

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**Question** According to the multiplication principle, it seems that the number of ways of choosing $a$ objects from $n$ should be equal to the number of ways of choosing $a-b$ objects from $n$ followed by $b$ objects from $n-(a-b)$ objects. If so, why doesn't
$$\binom{n}{a-b}\binom{n-(a-b)}{b}=\binom{n}{a}?$$