The digits 2, 2, 3, and 5 are randomly arranged to form a four digit number. What is the probability that the sum of the first and last digits is even?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{1}{6}$
(D) $\frac{1}{2}$
(E) $\frac{2}{3}$
The digits 2, 2, 3, and 5 are randomly arranged to form a four digit number. What is the probability that the sum of the first and last digits is even?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{1}{6}$
(D) $\frac{1}{2}$
(E) $\frac{2}{3}$

## Solution

The numbers that can be formed from the digits $2,2,3$, and 5, in ascending order, are 2235, 2253, $2325,2352,2523,2532,3225,3252,3522,5223,5232$, and 5322. These are twelve possibilities. For the sum of two digits to be even, both must be even or both must be odd. From the above list, the numbers for which the sum of the first and last digits is even, are 2352, 2532, 3225, and 5223. These are four possibilities. Thus, the probability of getting one of these numbers is $\frac{4}{12}$ or $\frac{1}{3}$.

ANSWER: (B)