a die is thrown four times what is the probability that each

a die is thrown four times, what is the probability that each number thrown is at least as high as all of the numbers that were thrown earlier

In how many ways can 11 men and 8 women stand in a row so that no two women stand next to each other

Solution

First consider the position of the men, then the position of the women.
In general first consider the the group with more people.

How many possible ways are there to arrange eight men in a row? It will be 11P11=11!=39916800
* M * M * M * M * M * M * M * M * M * M * M *

Hence, we need to find how many ways we can arrange 8 women in the 12 possible (as shown above) places,this is actually 12P8=19958400

Now applying the fundamental law of counting (precisely product rule), total number of possible arrangements satisfying both constraints is: 39916800 * 19958400 = 796675461120000