Analysis: One RMB in four different currencies in the question, namely 1, 2, 3 and 4 RMB in four different currencies, is selected as a group for combination.

1, single 1 RMB: 1 yuan, 2 yuan, 5 yuan, 10 RMB, ***4 currencies.

2. Two RMB combinations:

1+2=3 (yuan), 1+5=6 (yuan),1+10 =1(yuan), 2+5=7 (yuan)

***6 different combinations to get 6 different currencies.

3.3 RMB combinations:

1+2+5=8 yuan, 1+2+ 10= 13 yuan,1+5+10 =10.

***4 different combinations get 4 different currencies.

4.4 RMB combinations:1+2+5+10 =18 (RMB)

*** 1 different combinations get 1 different currencies.

Add the monetary values obtained from the above four combinations to 4+6+4+ 1= 15 (species), so there are *** 15 combinations.

This is a combinatorial problem in mathematics. Taking m(m≤n) elements from n (4 elements in the problem) as a group is called taking out the combination of m elements from n different elements. This problem of finding the number of combinations is called the combination problem.

The combination number can be directly calculated by the formula, which is:

For example: n=4, m is 1, 2, 3, 4 respectively.

When m is equal to 1, 2, 3 and 4 respectively, the results are: 4, 6, 4, 1. Add up the results to get 15. That is, there are 15 different combinations.

Extended data:

Combinatorial numbers have two properties.

1. Complementarity: that is, the number of combinations of m elements of n different elements = the number of combinations of (n-m) elements of n different elements.

For example, C(4, 1) = c (4,3), that is, the method of selecting1element from four elements is equal to the method of selecting three elements from four elements.

Provisions: c (n, 0) = 1 c (n, n) =1c (0,0) = 1.

2. Combinatorial identities

If m item is selected from n items, the following formula exists: c (n, m) = c (n, n-m) = c (n- 1, m- 1)+c (n- 1, m).

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