Thirty items are arranged in a 6-by-5 array as shown.

A1	A2	A3	A4	A5
A6	A7	A8	A9	A10
A11	A12	A13	A14	A15
A16	A17	A18	A19	A20
A21	A22	A23	A24	A25
A26	A27	A28	A29	A30

Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column.

• 200
• 760
• 1200
• 4560
• 7200

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

There are 10 (5 choose 3) ways to select the three columns in which the three items will appear. The row of the rightmost selected item can be chosen in any of six ways, the row of the leftmost selected item can then be chosen in any of five ways, and the row of the middle selected item can then be chosen in any of four ways. The answer is thus (10)(6)(5)(4) = 1200. Alternatively, there are 30 ways to select the first item. Because there are 10 squares in the row or column of the first selected item, there are 30 − 10 = 20 ways to select the second item. Because there are 18 squares in the rows or columns of the first and second selected items, there are 30 − 18 = 12 ways to select the third item. The number of permutations of three qualifying items is (30)(20)(12). The number of combinations is thus (30)(20)(12)/3! = 1200.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.