+1444 You are given the 4 x 4 grid below.
(a) Find the number of ways of placing 8 counters in the squares (at most one counter per square), so that each row contains exactly two counters.
(b) Find the number of ways of placing 12 counters in the squares (at most one counter per square), so that each column contains exactly three counters.
+73 A) The square has four columns and four rows. Each row and column has 4 squares. In each row, we have four squares to place the counters in each row, or 4 choose 2, which is \(\frac{(4\cdot3\cdot2\cdot1)}{(2\cdot1)}\). This equals 12. So, there are 12 ways to place 2 counters in each row. However, there are 4 rows, so we multiply by 4. 12 times 4 is \(\boxed{48}\).
+73 B) We can start the same way that we solved A). However, now we are dealing with the columns, and we have an extra counter. So, each column needs 3 counters. Instead of counting the number of ways we can place the counters, let's instead count the ways we can pick the empty square. We can pick each square in the column to be empty, so there are 4 ways. 4 ways to do this in each column, which is 4*4=16. So, the answer is \(\boxed{16}\).
