关于数列排列组合的问题!Judy and John are going out tonight for dinner and a movie.There are 3 restaurants they like and 4 movies they want to see.In how many ways can they choose a restaurant and a movie for this evening?A.3x4B.3+4C.3^4D.4

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关于数列排列组合的问题!Judy and John are going out tonight for dinner and a movie.There are 3 restaurants they like and 4 movies they want to see.In how many ways can they choose a restaurant and a movie for this evening?A.3x4B.3+4C.3^4D.4
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关于数列排列组合的问题!Judy and John are going out tonight for dinner and a movie.There are 3 restaurants they like and 4 movies they want to see.In how many ways can they choose a restaurant and a movie for this evening?A.3x4B.3+4C.3^4D.4
关于数列排列组合的问题!
Judy and John are going out tonight for dinner and a movie.There are 3 restaurants they like and 4 movies they want to see.In how many ways can they choose a restaurant and a movie for this evening?
A.3x4
B.3+4
C.3^4
D.4^3
E.2^3x2^4
Part 8 of 10 -
Question 8 of 10
1.0 Points
How many subsets of the set {A,B,C,D,E,F} contain the letter B but not the letter
A.4
B.8
C.12
D.16
E.32
Reset Selection
Part 9 of 10 -

关于数列排列组合的问题!Judy and John are going out tonight for dinner and a movie.There are 3 restaurants they like and 4 movies they want to see.In how many ways can they choose a restaurant and a movie for this evening?A.3x4B.3+4C.3^4D.4
第一题,两人要去吃饭并看电影,有3家饭店和4家电影院可选择,答案是3x4,
第二题求含B但不含C的子集合,
扣掉B和C还有A,D E,F4个
一个元素的也即只有,1种
两个元素的子集合就是B再加上4个中选1个,即C（4）（1）,组合的符号 前面4是下标,1是上标,有4种,
同理三个元素是C（4）（2）,6种,
4个元素是C（4）（3）4种
5个元素是C（4）（4）1种
所以共有16种

第一题选A 电影院C41 餐厅C31 C31XC41=4X3=12
第二题 相当于把B和C先弄出去 因为子集必然有B没有C 剩下4个 C41+C42+C43+C44=15 加上只有B的情况 一共16种 选D

第一个是A第二个是D

（1）乘法原理，A
（2）没有C，有B.
从A,D,E,F中选择，每个有2种选择，2^4=16，选D.

1.A 2.D 希望对你有帮助

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