MARY SAYS THAT THE GREATER THE NUMBER OF COUNTERS, THE GREATER THE NUMER OF DIFFERENT ARRAYS YOU CAN FORM. GIVE AN EXAMPLE THAT SHOWS THAT MARY IS WRONG.

If the "counters" are identical, there is only one possible array.

Does anyone know the answer I see postings for everything but the answer

To provide an example that shows Mary is wrong, we need to demonstrate a situation where increasing the number of counters does not result in a greater number of different arrays.

Let's consider a scenario with three different types of counters: red, blue, and green. We will explore two cases: one with three counters and another with four counters.

Case 1: Three Counters
Suppose we have three counters: one red, one blue, and one green. In this case, we can form different arrays by arranging these three counters in different orders. Here are all the possible arrays:

1) Red - Blue - Green
2) Red - Green - Blue
3) Blue - Red - Green
4) Blue - Green - Red
5) Green - Red - Blue
6) Green - Blue - Red

In total, we have six different arrays.

Case 2: Four Counters
Now, let's consider adding another counter, let's say an additional red counter. We now have four counters: two red, one blue, and one green. Here are all the possible arrays:

1) Red - Red - Blue - Green
2) Red - Red - Green - Blue
3) Red - Blue - Red - Green
4) Red - Blue - Green - Red
5) Red - Green - Red - Blue
6) Red - Green - Blue - Red

Wait a sec, these arrays are exactly the same as those generated in Case 1. Therefore, even though we added an extra counter, the number of different arrays remained the same at six.

This example disproves Mary's claim that the greater the number of counters, the greater the number of different arrays that can be formed. It's essential to carefully consider the properties and arrangements of the counters to determine the number of distinct arrays possible.