Let S be the set of five digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?
(1) 228
(2) 216
(3) 294
(4) 192

• 3c2*3p2*2c1*2!=74
  end (3) odd so=74/3=24*3=72
  72*3=216
• 9 years agoHelpfull: Yes(0) No(0)
• question may be number of digits formed
  each odd digit may be 3*2=6(Select two odd from three and arrange them)
  and arrange them is 3 ways
  so one odd digit must be even so select one odd place from three places is 3 and two select one even from two is 2 so number of ways is 3*2=6
  so remaining two places has two ways is 2*1=2 so 6*3*6*2=216
• 9 years agoHelpfull: Yes(0) No(0)