] Odd elements ={1,3,5,7,9} Total =5 Even elements ={2,4,6,8} Total =4 Case I 3 odd numbers numbers of ways =‌5C3=10 Case II 1 odd and 2 even numbers of ways =‌5C1×‌4C2 =5×6=30 Total favourable cases for A =10+30=40 And total ways for selecting 3 elements from 9 elements =‌9C3=84 ∴‌‌P(A)=‌