Binomial inversion problem with the wrong row

Binomial inversion problem with the wrong row

Common simple combinatorial identities:

1. \(C_n^m=C_n^{n-m}\)

2. \(C_n^m=C_n^{m-1}+C_{n-1}^{m-1}\)

3. \(\sum_{i=0}^{n}C_n^i=2^i\)

4. \(\sum_{i=0}^{n}(-1)^i*C_n^i=[n=0]\)

3.4 Proof: easy to prove the binomial theorem.

Order \ (. 1 X =, Y =. 1 \) , formula 3 can be obtained

Order \ (X =. 1, Y -1 = \) , the formula 4 can be obtained

Binomial inversion:

There are two functions f, g is assumed. Satisfies:
\ [F_n = \ sum_ I = {0}} ^ {n-G_i C_n ^ I * \]
then consider how to obtain trans \ (G_N \) on (F_n \) \ equation.
\ [G_n = \ sum_ {i = 0} ^ {n} [ni = 0] * C_n ^ i * g_i \\ g_n = \ sum_ {i = 0} ^ {n} \ sum_ {j = 0} ^ { ni} (- 1) ^ j * C_ {ni} ^ j * C_n ^ i * g_i \\ g_n = \ sum_ {i = 0} ^ {n} \ sum_ {j = 0} ^ {ni} (- 1 ) ^ j * C_ {n} ^ j * C_ {nj} ^ i * g_i \\ g_n = \ sum_ {j = 0} ^ {n} (- 1) ^ j * C_n ^ j \ sum_ {i = 0 } ^ {nj} C_ {nj } ^ i * g_i \\ g_n = \ sum_ {i = 0} ^ {n} (- 1) ^ i * C_n ^ i * f_ {ni} = \ sum_ {i = 0 } ^ {n} (- 1
) ^ {ni} * C_n ^ i * f_ {i} \] so obtained binomial inversion Conclusion:
\ [F_n = \ sum_ I = {0}} ^ {n-C_n ^ i * g_i \\ g_n = \
sum_ {i = 0} ^ {n} (- 1) ^ {ni} * C_n ^ i * f_ {i} \\ \] form is really beautiful!

Here binomial inversion to solve a classic problem!

Wrong row problem

Problem Description:

There \ (n \) personal number \ (1, ..., the n-\) , asking \ (n \) personal stand in a row the number of programs all the wrong location.

It is defined above the wrong: a first \ (I \) individuals do not stand in the position \ (I \) on.

Method 1: recursive

Set \ (f_n \) represents the answer, assuming that now takes into account the former \ (i \) individual programs, namely \ (F_i \) .

Consider the first \ (i \) individual stations circumstances:

Obviously the first \ (i \) person can not stand in the position \ (i \) , assuming he points to the location \ (k \) , apparently \ (k \ in [1, i-1] \) , then continue to consider \ (K \) stations.

①, \ (K \) stand to position i, then the remaining \ (i-2 \) individual original still pose a problem, the program number \ (I-2 F_ {} \) .

②, \ (k \) did not stand to position i, that is \ (k \) can not stand in the position \ (i \) , then the rest of \ (i-1 \) individuals continue to pose a problem of the original program number \ (F_i-1 \) .

Can be obtained \ (F \) of the recursion relation:
\ [F_1 = 0 \, \ \\ F_i = F_2. 1 = (. 1-I) * (F_ {I}. 1-I-2 + F_ {}) \ \ i≥3 \]

Method 2: Inversion binomial

Set \ (f_n \) represents \ (n \) individual stations just a few of the program, \ (G_N \) represents \ (n \) personal station number are wrong solution.

Easily obtained:
\ [= n-F_n F_n \\ = \ sum_ {0} I = I * ^ ^ nC_n G_i \!]
Direct inversion can be binomial:
\ [G_N = \ sum_ I = {0} ^ { n} (- 1) ^ {
ni} * C_n ^ i * f_ {i} \\ \] can also be introduced directly answer linear delivery.