Find the number of numbers that can formed using all the digits 1, 2, 3, 4, 3, 2, 1 only once so that the odd digits occupy the odd places only?
A.
4!/(2!)2
B.
7!/(2!)3
C.
1! 3! 5! 7!
D.
None of these

• 4 odd places and 4 odd numbers arrangement=4!/2!2!(two 1s & two 3s)
  3 even places and 3 even numbers arrangement=3!/2!(two 2s)
  ans=4!3!/(2!)3 (D)None
• 10 years agoHelpfull: Yes(19) No(1)
• ACCORDING TO ME....
  D)IS THE ANS BECAUSE odd places can be filled in 4!/2!*2! ways
  and even places can be filled in 3!/2!=3 ways hence total=4!*3/2!*2!
• 10 years agoHelpfull: Yes(6) No(0)