The number of passing routes of the ball in A's hand after N passes. Where n ≥ 2. Then f(2) = 5P 1 = 5, that is, (ABA

Automatic circuit analyzer

Automatic data acquisition (automatic data acquisition)

Atomic energy administration

AFA)f(3) = 5P2 = 20, i.e. (ABCA)

ABDA

ABEA

ABFA·ACBA

ACDA

ACEA

ACFA·ADBA

ADCA

ADEA

ADFA·AEBA

AECA

AEDA

AEFA· afba

AFCA

AFDA

AFEA) f(4) There are two situations: situation 1: Add BA after the last two students in f(2).

Canada

Daily Allowance (daily allowance)

EA or FA. *** 5 f(2) species. Case 2: Add a classmate between the last two classmates in case f(3).

There are four possibilities.

*** 4 f(3) species. So f (4) = 5f (2)+4f (3) = 5 * 5+4 * 20 =105.

F (5) = 5f (3)+4f (4) = 5 * 20+4 *105 = 520 * * * There are 520 passing routes. For similar topics, please refer to the free classics of the first generation of mathematicians:. Qid = 7010110501309 here shows another highly abstract analytical thinking.

Think carefully.

1 time = 5C 1 time, 4C 1 time, or 1C 1 time, the second transmission shall not be made from the hand of a. If it is in A's hand for the third time, it will come out of A's hand for the fourth time. If it is in A's hand for the fifth time, it will be in A's hand.

Then 4c 1 and 1c 1 If the third time is not in A's hands, then the fourth time and the fifth time are not in A's hands, only in A's hands.

Then 3c 1 and 1c 1 * * have a passing route = 5c1* [4c1* 3c1*1*. = 5(4*3* 1+ 1*4* 1) = 5( 12+4) = 80?