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Next we replace each of x1, x2, and x3 by x in our two expressions for M. This results in
 

(1 + x)3 = 1 + 3x + 3x2 + x3.

Thus we see thatfor k = 0, 1, 2, 3 is the number of ways of choosing a subset of k elements from a set S of 3 elements. Similarly, one can see that the number of ways of choosing k elements from a set of n elements is 
For example, the set {1, 2, 3, 4, 5} with 5 elements has subsets having 3 elements. Since

it is not too difficult to write out all ten of these subsets as
 

{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5},
{1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}.

If we drop the braces enclosing the elements of each subset, the resulting sequence is said to be a combination of 3 things chosen from the set {1, 2, 3, 4, 5}. Thus the ten combinations of 3 things chosen from this set of 5 objects are
 
 

1, 2, 3; 1, 2, 4; 1, 2, 5; 1, 3, 4; 1, 3, 5;
1, 4, 5; 2, 3, 4; 2, 3, 5; 2, 4, 5; 3, 4, 5.
 

Note that changing the order in which the objects of a combination are written does not change the combination. For example, 1, 2, 4 is the same combination as 1, 4, 2.
 

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Tuesday, May 12, 1998