Counting nCr = n!/r!(n-r)!=nC(n-r) This equation reflects the fact that selecting r items is same as selecting n-r items in forming a combination from n items.

Compare two problems Number of strings of 0,1, and length n 2 n Number of strings of 0,1, length n, and r 0s nCr Consider the set {1,2,…,n}, the positions of the digits in the strings. There are nCr ways to choose r positions to assign 0.

Theorem 2. Suppose k selections are to be made from n items without regards to order and that repeats are allowed, assuming at least k copies of each of the n items. The number of ways these selections can be made is (n+k-1) C k.

Example. Choose 3 CDs from 10 (allow repeats). Consider the CDs one by one. 1 means select one copy and 2 means no (more) copies from the current choice. E.g., represents selecting CD 4,6, represents CD 1,two copies 2 Conversely, any selection can be represented by such a sequence. The string has three 1’s and ten 2’s. The last digit is always 2. The first 3+9 digits can be either 1 or 2. Therefore, total number 12 C 3.

This problem is equivalent to find the number of ways to put r identical items (or tokens, to give another name) into n different boxes. This is easy to see. Associate an object with a box. Choosing a copy of a certain object is associated with putting a token into the associated box. E.g., selecting CD 3, two copies of CD 6 is associated with putting one token in box 3 and two tokens in box 6.

Let 0 denote a token. A string of r 0’s and (n-1) 1’s models one assignment of r tokens into n boxes. Consider the same example as before, with r=3 and n= represents putting one token each in box 4,6 and represents putting one token in box 1, two in box 2 Therefore, total number 12 C 3.