Patrick has a locker with a 3 number combination. The lock contains numbers from 1 thru 30. No number may be used more than once. Patrick is superstitious and never uses numbers that end with an odd number. He also only uses 2 digit numbers. How many possible 3 digit combinations could he use?  Choose the correct answer.
A) 110
B) 440
C) 550
D) 880
E) 990
Permutation calc is: n P r = n! / (n-r)! 11 P 3 = 11!/(11-3)! = 11x10x9x(8x7x6x5x4x3x2x1)/8!=11x10x9=990 Correct answer is 990 (990)
<result>E</result>
Patrick has a locker with a 3 number combination. The lock contains numbers from 1 thru 30. No number may be used more than once. Patrick is superstitious and never uses numbers that end with an odd number. He also only uses 2 digit numbers. How many possible 3 digit combinations could he use?