A) 18
B) 28
C) 6
D) 27
Correct Answer: A
Solution :
[a] There are 4 odd digits (1, 1, 3, 3) and 4 odd place (first, third, fifth and seventh). At these places the odd digits can be arranged in \[\frac{4!}{2!2!}\]=6 ways Then at the remaining 3 places, the remaining three digits (2, 2, 4) can be arranged in \[\frac{3!}{2!}=3\] ways \[\therefore \] The required of number of numbers \[=6\times 3=18\]
You need to login to perform this action.
You will be redirected in
3 sec
