Virginia writes down the value of the \(100\) products \(a \times b\), where \(a\) and \(b\) represent integers from \(0\) to \(9\), inclusive. How many distinct numbers does Virginia write?
Virginia writes down the value of 100 products of integers from 0 to 9. For each product, she has 10 choices for the first integer and 10 choices for the second integer, for a total of 10⋅10=100 possibilities. However, some of these products will be the same.
For example, both 2⋅5 and 5⋅2 are equal to 10, so these two products are counted twice. To avoid this overcounting, we can use the following formula to count the number of distinct products:
where n is the number of distinct elements in the set and r is the number of elements in each product. In this case, we have n=10 distinct elements (the digits 0 to 9) and r=2 elements in each product. So, the number of distinct products is:
Using the combination formula, we can calculate that there are (211)=55 distinct products. Therefore, Virginia writes down 55 distinct numbers.