Detailed explanation of this problem for those who are beginning P&C
First Day Choice: 3C1*3C1*2C1 ( One from three shirts, one from three pants, one from two pair of shoes)
Second Day Choice : 2C1 * 2C1* 1C1 ( One from two shirts(one shirt is already selected) , one from two pants(one pant is already selected), one from one pair of shoes (same shoe for the second day)
Third Day Choice : 1C1 *1C1*1C1 (one shirt , pant and one pair of shoe(same shoe))
So the combination for three days is
(3C1*3C1*2C1) *(2C1 * 2C1* 1C1)* (1C1 *1C1*1C1) -----> 1
Total possible combination for three days without constraint(i.e selecting one on each day) is
(3C1*3C1*2C1) *(3C1 * 3C1* 2C1)* (3C1 *3C1*2C1) ------> 2
Probability = favorable Events / Total Events
Therefore dividing 1/2 we get (1/3)^4 , Answer C