bharat bajaj
Last Activity: 10 Years ago
O E O E O E O E O
There are 9positions, 5 odd and 4 even.
Also, there are 4 odddigitsand 5 evendigits.
No. of ways in which odddigitscan occupy evenpositions= 4! / (2! 2!)
No. of ways in which evendigitscan occupy oddpositions= 5! / (3! 2!)
Hence, required no. of ways = (4! 5!) / (2! 2! 3! 2!) = 120
THanks
Bharat Bajaj
askiitians faculty