Number of methods =4! -C(4， 1)*3！ +C(4，2)*2！ -C(4，3)* 1！ +C(4，4)*0！ =9 species

It is also possible to discuss this situation:

One of the three positions behind row A has C(3, 1)=3 methods.

Then discuss, for example, A ranks in the third position: then if C ranks in the position of A, then B and D are left, and only 1 is the sorting method; If C is not in the position of A, then one of B and D will be in the position of A. Once it is determined whether it is B or D, the sorting method is determined. There are two kinds at this time.

So after the position of A is determined, there are 1+2=3 methods, so a * * * has 3*3=9 methods.

The second problem is to send in the air:

First, the two old people are regarded as a whole. Then there are five kinds of A (5,5), four of which are empty, that is, A (5,5) * 4 * 2 = 960.

The third question:

* * * has 22 children.

Process:

200-2= 198

198= 1 1*2*3*3=22*9

So, the answer is 22.

In other words, 22 counts are exactly one lap.

The fourth question:

100g After pouring out 40g of salt water, it is 60g 80% 80% salt water with a salt content of 60 * 0.8 = 48g. Adding 40g of water is 48÷ 100=48%. After pouring out 40 grams of water, it is now 60 grams of 48% saline, so the final concentration is (60

The fifth question:

Because the first pile is 1/4 more than the second pile, the first pile is 5/4 times that of the second pile, 7/8 times that of the second pile and 7/8 times that of the first pile after transporting 3.6 tons from the first pile.

The second coal pile: 3.6/(5/4-7/8) = 9.6 (ton)

First coal pile: 9.6 * 5/4 = 12 (ton)

One last question:

20 species