how to approach SQL query that counts various combination of boolean flags?

I have this table structured like the below, and I want to be able to count all the predefined scenarios I'm searching for -

The real data is obviously much more complicated but if for simplicity sake I use a clear data here. I'm struggling to find a smart way of doing that instead of having aggregated table per each "stat".

here's the script for generating the example:

declare @persons table(
person_id INT,
country_id INT,
is_adult BIT,
is_fat BIT,
is_tall BIT,
is_bald BIT)

declare @stats table(
    stat_name varchar(max),
    persons_count INT
)

insert @persons
values
    (1000, 12, 1, 0, 1, 0),
    (1001, 15, 1, 0, 0, 1),
    (1002, 12, 0, 1, 1, 0),
    (1003, 20, 1, 0, 0, 0),
    (1004, 15, 0, 1, 0, 1)

insert @stats
values
    ('adults, fat, tall', 0),
    ('adults, fat, tall, bald', 0),
    ('fat, bald', 0),
    ('tall', 0)

Thx

You can create a bit mask for each value and then match on that mask from your stat table:

DECLARE @persons TABLE(
person_id INT,
country_id INT,
is_adult BIT,
is_fat BIT,
is_tall BIT,
is_bald BIT
, agg AS is_adult + 2 * is_fat + 4 * is_tall + 8 * is_bald
)

DECLARE @stats TABLE(
    stat_name varchar(max),
    persons_count INT, agg int
)

INSERT @persons
VALUES
    (1000, 12, 1, 0, 1, 0),
    (1001, 15, 1, 0, 0, 1),
    (1002, 12, 0, 1, 1, 0),
    (1003, 20, 1, 0, 0, 0),
    (1004, 15, 0, 1, 0, 1)

INSERT @stats
VALUES
    ('adults, fat, tall', 0, 1 + 2 + 4),
    ('adults, fat, tall, bald', 0, 1 + 2 + 4 + 8),
    ('fat, bald', 0, 2 + 8),
    ('tall', 0, 4),
    ('adult,tall', 0, 1 + 4)

SELECT	s.stat_name, count(*)
FROM	@persons p
INNER JOIN @stats s
	ON	s.agg & p.agg = s.agg
GROUP BY s.stat_name

i added agg column which is a 4 bit mask of 1 or 0es depending on your is_trait field.
It follows the 2 ** 0, 2 ** 1, 2 ** 2, ... 2 ** N pattern for each new field.
Binary-wise it looks like: 1101 for adult, slim, tall and bald (mask goes from the right).

For the stats, you create same mask of the fields you wanna match.

Finally for the join, you do a bit & (and) operator and check so your result matched. For example: 0111 & 1001 will return bits where both bits in same position is 1, ie: 0001

EDIT: Great username