KCET 2019 Math Solved Paper

(C)
Given digits are 1, 2, 3, 4, 5, 6, 7.
Two even digits can be selected in 3C2
Two odd digits can be selected in 4C2 ways.
These selected 4 digits can be arranged in 4! ways.
∴ Total number of ways = 4C2 3C2 4!
= 6 × 3 × 24
= 18 × 24
= 432