从 9 名男性和 12 名女性中选出一个由 5 名男性和 3 名女性组成的委员会，有几种方法？
组合是从集合中选择的项目，因此选择的顺序无关紧要。基本上，它可以被认为是无序排列。 k-combination 可以被称为 n 个事物的组合，一次取 k 个，不重复。例如，我们有三个字母 A、B 和 C，我们需要从中选择任意两个字母，得到可能的组合：AB、BC 和 AC。在组合的情况下，项目的选择顺序无关紧要，即 AB 本质上等同于 BA。
nCr = n!/(n – r)!r!
Where,
n – total number of objects from which to choose data
r – the number of chosen objects
由 5 男 3 女组成的委员会，可以从 9 男 12 女中选出多少种方式
解决方案：
Since, combinations are referred to using the following formula:
nCr = n!/(n-r)!r!
Since we need to choose, 5 men out of 9 men,
For Men = 9C5
Since, we need to choose, 3 women out of 12 women,
For Women = 12C3
Now,
Choose 5 men out of 9 men
Therefore,
nCr = n!/(n – r)!r!
9C5 = 9!/(9 – 5)!5!
9C5 = 9!/4!5!
9C5 = 9 × 8 × 7 × 6 × 5!/4 × 3 × 2 × 1 × 5!
9C5 = 3024 × 5!/24 × 5!
Simplifying Further,
9C5 = 3024/24
9C5 = 126
Thus,
We can choose 5 men out of 9 men in 126 ways
Now, further calculating for women,
Choose 3 women out of 12 women
Therefore,
nCr = n!/(n – r)!r!
12C3 = 12!/(12 – 3)!3!
12C3 = 12!/9!3!
12C3 = 12 × 11 × 10 × 9!/9! × 3 × 2 × 1
12C3 = 1320 × 9!/6 × 9!
Simplifying Further
12C3 = 1320/6
12C3 = 220
Thus,
Choose 3 women out of 12 women in 220 ways
Therefore,
The committee can be chosen in 126 × 220 = 27720 ways.
示例问题
问题1：说明一些排列组合的例子？
解决方案：
Permutation CombinationArranging people, digits, numbers, alphabets, letters, and colours Selection of menu, food and team. Picking a team captain, pitcher from a group Picking three team members from a group. Picking two favourite colours, in order, from a colour brochure. Picking two colours from a colour brochure. Picking first, second and third place winners. Picking any three winners.
问题 2：求 n = 4 和 r = 2 的排列组合？
解决方案：
Permutation:
nPr = n!/(n – r)!
4P2 = 4!/(4 – 2)!
4P2 = 4!/2!
4P2 = 4 × 3 × 2!/2!
Simplifying
4P2 = 4 × 3
4P2 = 12
Combination:
nCr = n!/(n – r)!r!
4C2 = 4!/(4 – 2)!2!
4C2 = 4!/2!2!
4C2 = 4 × 3 × 2!/2 × 1 × 2!
Simplifying
4C2 = 4 × 3/ 2
4C2 = 12/2
4C2 = 6
问题3：一束有6朵玫瑰和8朵百合花的花，可以从8朵玫瑰和10朵百合中选择多少种？
解决方案：
Here choose flowers for bouquet
Finding how rose can be chosen
6 rose from 8 rose = 8C6 = 8!/(8-6)!6!
= 8!/2!6!
= 8 × 7 × 6!/2 × 1 × 6!
Simplifying
= 56/2
= 28
8C6 = 28
Roses can be chosen in 28 ways
Finding how lily can be chosen
8 lily from 10 lily = 10C8 = 10!/(10 – 8)!8!
= 10!/2!8!
= 10 × 9 × 8!/2 × 1 × 8!
Simplifying
= 90/2
= 45
10C8 = 45
Lily can be chosen in 45 ways
The flowers for bouquet can be chosen in 1260 ways.
问题 4：评估 C(22, 20)。 C(n, r) = n!/(n – r)!r!
解决方案：
Here,
n = 22
r = 20
Putting the values of n and r in C(n, r) = n!/(n – r)!r!
C(22, 20) = 22!/(22 – 20)!20!
C(22, 20) = 22!/(2)!20!
C(22, 20) = 22 × 21 × 20!/2! × 20!
C(22, 20) = 22 × 21/ 2 × 1
C(22, 20) = 462/2
C(22, 20) = 231
Therefore,
C(22, 20). C(n,r)=n!/(n-r)!r! = 231
问题 5：找出 C(30, 28)。 C(n, r) = n!/(n – r)!r!
解决方案：
Here,
n = 30
r = 28
Putting the values of n and r in C(n, r) = n!/(n – r)!r!
We get,
C(30, 28) = 30!/(30 – 28)!28!
C(30, 28) = 30!/(2)!28!
C(30, 28) = 30 × 29 × 28!/2! × 28!
C(30, 28) = 30 × 29/ 2
C(30, 28) = 870/2
C(30, 28) = 435
Therefore,
C(30, 28). C(n,r)=n!/(n – r)!r! = 435