Ask for explanation Given six numbers 0 1 2 3 5 9, how many natural numbers can be formed without repetition?

one thousand six hundred and thirty-one

How about permutation C

C(n,m)=n!/m!(n-m)!
x!=x(x-1)(x-2)…… 3*2*1

How to prove ∑ [i = 0, n] C (m, I) C (n-m, n-i) = C (n, n) by permutation and combination

By using (a + b) ^ n = (a + b) ^ m x (a + b) ^ (n-m), the coefficients of a ^ n x B ^ (N-N) on both sides of the equation are investigated, and the equation is obtained

Permutation and combination C M N M = 10 N = 0 this is equal to several ah
The probability C is a 0 above and a 10 below

C Mn algorithm is a m n divided by N level, this is more special, the answer is 1

Permutation and combination problems a (m, n) and C (m, n) * a (n,

A(M,N)＝C(M,N)*A(N,N)
A (m, n) is the permutation number of n elements selected from m different elements
C (m, n) is the combination number of n elements selected from m different elements (regardless of the order of elements)
A (n, n) is the total permutation number of n elements

Detailed algorithm of permutation and combination (a, c)

A(a,b)=a!/b!
C(a,b)=a!/[b!*(a-b)!]

High school permutation and combination problems do not need to calculate
There are three bags marked as a, B and C. There is a red ball in bag a, two different white balls in bag B and three different yellow balls in bag C. now take out two balls from them. If the two balls are different in color, how many ways can we take them
Is it a way to take out the red ball first and then the white ball and then the red ball?

It's one, but you should pay attention to the same color ball is also different, so if you take the first white ball and then the red one, and you take the red one and then the second white one, it's not one

For high school permutation and combination of a and C algorithm, it is best to use specific numbers to illustrate the following

A x = x * (x-1) * (X-2)... * (X-Y + 1) Y C x = x * (x-1) * (x-1)... * (X-Y + 1) / Y / (Y-1) / (Y-2)... / 1 y, for example, a 4 = 5 * 4 * 3 * 2 = 120 5 C 4 = 5 * 4 * 3 * 2 / 4 / 3 / 2 / 1 = 5

What is the difference between a and C in permutation and combination? What are their respective algorithms?
Please~

Ordered arrangement disordered combination
A（m,n）=n!/(n-m)!
0!=1
C（m,n）=n!/((n-m)!*m!)
c(n,m)=c(n,n-m)

Permutation and combination calculation: C (0) [3] + C (1) [4] + C (2) [5] +... + C (17) [20]
Calculation: C (0) [3C (1) [4] + C (2) [5] +... + C (17) [20]
PS: superscripts in parentheses and subscripts in parentheses

Original formula = 3C3 + 3C4 + +3C20
=4C4+3C4+…… +3C20
kC（n+1）=（k-1）Cn+kCn
SO
4C4+3C4=4C5
4C5+3C5=4C6
The final formula is 4c21

Ask for explanation Given six numbers 0 1 2 3 5 9, how many natural numbers can be formed without repetition?