In the $3 \times 3$ table shown, the numbers 1,2 and 3 are placed so that each number occurs only once in each row and only once in each column. The value of $X+Y$ is
(A) 3
(B) 2
(C) 5
(D) 6
(E) 4
Consider the value of $U$ in the diagram shown.

Since a 3 already occurs in the second row, then $U$ cannot equal 3 (each of the numbers 1,2,3, occur only once in each row).

Since a 1 already occurs in the third column, then $U$ cannot equal 1

(each of the numbers 1, 2, 3, occur only once in each column).

Since $U$ cannot equal 3 or 1 , then $U=2$.

Therefore, a 2 and a 3 already occur in the second row and so $X=1$.

|  |  | 1 |
| :--- | :--- | :--- |
| 3 | $X$ | $U$ |
|  |  | $Y$ |

At this point, a 2 and a 1 already occur in the third column and so $Y=3$.

The value of $X+Y=1+3=4$.

ANSwer: (E)