请教一道数学题-排列组合

piscesu · 发表于 2019-5-16 10:42:50

Set AA consists of kk distinct numbers. If nn numbers are selected from the set one-by-one, where n≤kn≤k, what is the probability that numbers will be selected in ascending order?

(1) Set AA consists of 12 even consecutive integers.

(2) n=5

OA=B

实在不明白为什么是B以下是official explanation
We should understand the following two things:

1. The probability of selecting any nn numbers from the set is the same. Why should any subset of nn numbers have higher or lower probability of being selected than some other subset of nn numbers? Probability doesn't favor any particular subset.

2. Now, consider that the subset selected is {x1, x2, ..., xn}{x1, x2, ..., xn}, where x1<x2<...<xnx1<x2<...<xn. We can select this subset of numbers in n!n! # of ways and out of these n!n! ways only one, namely {x1, x2, ..., xn}{x1, x2, ..., xn} will be in ascending order. So 1 out of n!. P=1n!P=1n!.

Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (nn) we are selecting from set AA.

Answer: B

hedyzou · 发表于 2019-5-16 10:51:59

这是哪的题啊

一颗野生的蛋蛋 · 发表于 2019-5-17 00:47:34

这题好难啊……个人觉得考试的时候不会出这么难的题qwq
看完了official explanation, 通俗的解释一下思路：由于题干中说集合A中的元素都是distinct numbers，因此，任意从集合A中抽出n个元素，这n个元素有且只有1种升序排列，而从集合A中抽取到这特定的n个元素的方法有n!种，因此抽到升序的概率就是1/n!，跟A中元素的个数无关。
或者这样解释：从k个元素中抽取n个元素，共有C（n,k）种方法，每种抽法对应1种升序排列，即共有C(n,k)种可能的升序排列；而考虑顺序，从k个元素中抽取n个元素，共有A(n,k)种抽法，因此，升序的概率就是C(n,k) / A(n,k)，即1/n!

piscesu · 发表于 2019-5-17 01:51:28

hedyzou 发表于 2019-5-16 10:51
这是哪的题啊

gmatclub...

piscesu · 发表于 2019-5-17 01:53:10

一颗野生的蛋蛋发表于 2019-5-17 00:47
这题好难啊……个人觉得考试的时候不会出这么难的题qwq
看完了official explanation, 通俗的解释一下思路： ...

懂了！

请教一道数学题-排列组合

所属分类: GMAT考试

正在浏览此版块的会员 ()