Steps to reproduce this combinatorial result?

Group A: Q, w, e, r

Group B: a, s, d

Group C: Z, x, c, v, b, n

Group D: y, u, i

Picking one item from each group to make a four course dinner will give us 216 combinations. But suppose the waiter allows to skip a group completely, then what will be the number of combinations? The solution to this problem has been described as 5 * 4 * 7 * 3 = 560. Can someone explain how this is possible? I thought by omitting a group completely, the number would decrease.

Under the new rule you can still pick one item from each group, so you can still order every dinner that you could have ordered under the original rule. However, you now have new possibilities: for instance, you can choose $$w$$ from Group A, $$x$$ from Group C, $$y$$ from Group D, and nothing from Group B. The new rule adds a whole bunch of three course dinners to that weren’t available under the old rule. It also allows a bunch of two course and one course dinners: you could, for instance, just order $$u$$ from Group D and nothing else.
As for the actual mathematics, in effect each group has a new choice, none of these. From Group B, for instance, you can choose $$a$$, $$s$$, $$d$$, or nothing at all. Thus, you now have $$4$$ choices in Group B instead of $$3$$. The same goes for each of the other groups: in effect Group A now has $$5$$ choices instead of $$4$$, Group C has $$7$$ instead of $$6$$, and Group D has $$4$$ instead of $$3$$. Thus, there are now $$5\cdot 4\cdot7\cdot4=560$$ ways to choose one item from each group. Note, though, that one of these is to choose the none option from every group, in which case one really isn’t ordering a meal at all, so I’d be inclined to say that the number of possible dinners is actually $$560-1=559$$.