How many ways to arrange deck of cards (4 suits from 6 to ace = 36 cards) so that no king comes before any diamond card are there

Question

So we have 36 cards and we need to find how many ways are there to arrange all the cards so that no king comes before any diamond card.

I have totally no idea how to solve it, I don't understand how to calculate the ways to place all 4 kings to the desired position

Answer 1

There is one king of diamonds, $3$ other kings and there are $8$ non-king diamonds.

There are $\binom{36}{12}$ ways to pick the positions for these important cards.

We ignore all other cards and assume only the important cards exist.

Notice that if we look at the important cards there are $8!$ ways to order the non-king diamonds (they have to go at the beginning).

There are $3!$ ways to order the non-diamond kings (they go at the end of the important cards).

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We can now order all of the other cards in $24!$ ways.

Therefore the answer is $\binom{36}{12}8!\times3!\times24!$