A) 180
B) 7560
C) 60
D) 16
Correct Answer: C
Solution :
4 odd digits in 4 even places can be arranged in 4! Ways. But 2 alike \[(3,3),2\]alike \[(5,5)\]are there \[\therefore \]Number of ways \[=\frac{4!}{2!2!}=\frac{24}{2}=6\] ?(i) Remaining 5 even digits (2 alike, 3 alike) in 5 places can be arranged as above in \[\frac{5!}{2!3!}=\frac{120}{2\times 6}=10\,\text{ways}\] ?(2) By fundamental theorem real number \[=6\times 10=60\]You need to login to perform this action.
You will be redirected in
3 sec