题:7人站成一排,其中甲不在排头或不在排尾的排法共有多少种? 解:因甲不在排头或不在排尾,故只能在中间五个位置之一。第一步先排甲有P(1/5)种,第二步再将剩余的六个位置排除甲外的6人有P_6~d种,据乘法原理所求排法种数为P_5~1P_6~6=3600注意:甲不在排头且不在排尾的排法,也是P_6~1P_6~6种,这岂不是“或”与“且”一样吗? Question: 7 people stand in a row, in which A is not in the first row or in the end of the row of a total of how many? Solution: because A is not in the first row or not in the end, it can only be in one of the middle five positions. In the first step, there are P(1/5) species in the first row, and 6 people in the second step to exclude the remaining six locations have P_6~d species. According to the principle of multiplication, the number of species to be excluded is P_5~1P_6. ~6=3600 Note: A is not the first row and is not in the row. It is also P_6~1P_6~6. Isn’t this the same as “and”?