从 10 个女孩和 12 个男孩中选出一组 4 个女孩和 7 个男孩有多少种方法？
在数学中，排列被称为排列一个集合的过程，其中一个集合的所有成员被排列成一些系列或顺序。如果集合已经排列，则排列的过程称为对其组件的重新排列。几乎所有数学领域都以或多或少的重要方式发生排列。当考虑某些有限集上的不同命令时，它们经常出现。
置换公式
在排列中，从 n 个事物的集合中选择 r 个事物，没有任何替换。在这个选择的顺序。
npr = (n!)/(n – r)!
n = set size, the total number of items in the set
r = subset size, the number of items to be selected from the set
组合
组合是从组中选择项目的行为，这样（不像排列）选择的顺序无关紧要。在较小的情况下，可以计算组合的数量。组合是指一次取k个不重复的n个事物的并集。组合可以以任意顺序选择项目。对于那些允许重复出现的组合，经常使用术语 k-selection 或 k-combination with replication。
组合配方
组合 r 个事物是从一组 n 个事物中选择的，其中选择的顺序无关紧要
nCr =n! ⁄ ((n-r)! r!)
Here, n = Number of items in set
r = Number of items selected from the set
从 10 个女孩和 12 个男孩中选出一组 4 个女孩和 7 个男孩有多少种方法？
解决方案：
Here we use the formula for selecting r things from n different things.
Complete step-by-step answer:
The number of ways of selecting r things from n different things is given by nCrand
the value of nCr is given as nCr = n! ⁄ r!(n−r)! .
Here sequence doesn’t matter
The formula of selecting 4 girls from 10 girls: 10C4
= 10! ⁄ 4!(10-4)!
= 10! ⁄ 4!6!
=10 × 9 × 8 × 7 × 6!⁄ 4 × 3 × 2 × 1 × 6!
= 10 × 9 × 8 × 7⁄4 × 3 × 2
= 10 × 9 × 8 × 7⁄8 × 3
= 10 × 9 × 7⁄3
= 10 × 3 × 76
= 210
The formula of selecting 7 boys from 12 boys: 12C7
= 12! ⁄ 7!(12-7)!
= 12! ⁄ 7!5!
= 12 × 11 × 10 × 9 × 8 × 7! ⁄ 7! × 5 × 4 × 3 × 2 × 1
= 12 × 11 × 10 × 9 × 8 ⁄ 10 × 12
= 11 × 9 × 8
= 792
No of ways to make the team of 4 girls and 7 boys = 10C4 × 12C7
= 210 × 792
= 1,66,320 ways
类似问题
问题 1：一组由 4 个女孩和 7 个男孩组成。如果团队至少有一个男孩和一个女孩，有多少种方法可以选择一个 5 人的团队。
解决方案：
From a group of 4 girls and 7 boys choose 5 members, such that
there is at least one boy and one girl in the team.
So, first let’s select one boy and one girl. The number of ways of selecting 5 number
is given as 7C1 × 4C1 = 7×4=28.
Now, the abide three members of the team can be either girls or boys . So, we have to
select three members from a group of nine persons i.e., three girls and six boys.
So, number of ways of selection is given as 9C3=9! ⁄ 3!×6!=9×8×7 ⁄ 6=84
So, the number of ways of choosing a team of 5 members from a group of 4 girls and
7 boys, such that there is at least one boy and one girl in the team is given as
7C1×4C1×9C3 = 28×84 = 2352
Hence, there are 2352 ways of choosing a team of 5 members from a group of 4 girls and
7 boys, such that there is at least one boy and one girl in the team.
问题 2：一组由 4 个女孩和 7 个男孩组成。如果团队没有女孩，有多少种方法可以选择 5 名成员的团队。
解决方案：
We have to choose a team of 5 members from a group of 4
girls and 7 boys, such that there are no girls in the team, i.e. all the members should
be boys.
So, we have to select 5 boys from a group of 7 boys.
So, number of ways =7C5
=7! ⁄ 5!(7−5)!
=7! ⁄ 5!×2!
Now, we know, we can write n! as n(n−1)!=n(n−1)(n−2)! . So, 7!=7×6×5!
So, the number of ways =7×6×5! ⁄ 5!×2!=7×6 ⁄ 2=21.
Therefore , there are 21 ways of choosing a team of 5 members from a group of 4 girls
and 7 boys, in such a way that there are no girls in the team.
问题3：4个女孩和1个男孩有多少种方式，可以从4个女孩和7个男孩中选出组队？
解决方案：
Here we use the formula for selecting r things from n different things.
Complete step-by-step answer:
The number of ways of selecting r things from n different things is given by nC rand
the value of nCr is given as nCr=n! ⁄ r!(n−r)! .
Here sequence doesn’t matter
The formula of selecting 4 girls from 4 girls :- 4C4
= 4! ⁄ 4!(4-4)!
= 4! ⁄ 4! 0!
=1
The formula of selecting 1 boys from 7 boys :- 7C1
= 7! ⁄ 1!(7-1)!
=7! ⁄ 1! 6!
= 7 × 6! / 6!
= 7
No of ways to make the team of 4 girls and 1 boys = 4C4 × 7C1
= 1 × 7
= 7 ways
问题4：3个女孩和2个男孩有多少种方式，可以从4个女孩和7个男孩中选出组队？
解决方案：
Here we use the formula for selecting r things from n different things.
Complete step-by-step answer:
The number of ways of selecting r things from n different things is given by nCr and
the value of nCr is given as nCr=n! ⁄ r!(n−r)! .
Here sequence doesn’t matter
The formula of selecting 3 girls from 4 girls :- 4C3
= 4! ⁄ 3!(4-3)!
= 4! ⁄ 3! 1!
=4 × 3! / 3!
= 4
The formula of selecting 2 boys from 7 boys :- 7C2
= 7! ⁄ 2!(7-2)!
=7! ⁄ 2! 5!
= 7 × 6 × 5! / 2! 5!
= 7 × 6 / 2
= 7 × 3
=21
No of ways to make the team of 3 girls and 2 boys = 4C3 × 7C2
= 4 × 21
= 84 ways