The answer is $\frac{52!}{(5!\times 5!\times 5!\times 5!\times 32!)} = 1.4782628e+24$
This reminds me of that if I have to select 7 objects from 3 types with infinite supply:
oo|oo|ooo
I know that it's no necessary to use the bar in this question because each person has the same numbers of the cards, but I still want to try to add 4 bars.
That is:
5 cards for person1 | 5 cards for person2 | 5 cards for person3 | 5 cards for person4 | undealt cars
so the equation is: $\frac{56!}{(5!\times 5!\times 5!\times 5!\times 32!\times 4!)}=5.4295116e+29$
What's wrong with my idea?
Stars and bars is useful for cases where the number of cards each person gets is not fixed and the cards are considered identical. You don't care about which cards each person gets, just how many. You then place the bars to show how many cards each person gets. For example, if you were distributing $52$ cards to five people with no restriction that any person get any cards, you would imagine adding one to each hand and distributing $57$ cards to five people with each one getting at least one. You would imagine $57$ stars and place four bars to separate the cards that each one gets. There are $56$ places for bars, so there are $_{56}C_4$ ways to do this. In this case all the cards get distributed to somebody. This is very different from the problem you are working.