1: Set five pairs of shoes as follows.
A 1a2, b 1b2, c 1c2, d 1d2, e 1e2( 12 represents the left and right feet respectively).
Discuss the case of taking a pair and adding the same one:
(1) take a 1a2, c (5,1); Then take b 1, c (8,1); Then take c 1, C(6, 1)
The result is a 1a2b 1c 1, and the combination number is c (5, 1) * c (8, 1) * c (6, 1).
(2) take a 1a2, c (5,1); Then take c 1, c (8,1); Then take b 1, C(6, 1).
The result is a 1a2c 1b 1, and the combination number is c (5, 1) * c (8, 1) * c (6, 1).
It can be seen that although the two methods are different, the result is the same, that is, the same combination, and the combination is repeated.
Dividing by a (2,2) is the case of eliminating duplication.
2. Set two TV models as A, B, C, D, 1, 2, 3, 4, 5.
( 1)AB 1、AB2……C(4，2)*C(5， 1)
(2)A 12、A 13……C(5，2)*C(4， 1)
It can be seen that there will be no repetition and there is no need to divide by a (2,2).
The difference between the first question and the second question is that:
The first questions b 1 and c 1 are calculated independently, that is, C(8, 1)*C(6, 1), and b1c1b will appear.
The second question, 1 2, is calculated as a whole, that is, c (5,2). The combined calculation avoids the situation that 12=2 1.
Not only the arrangement will be repeated, but also the combination will be repeated. Be sure to pay attention.