Question -
Answer -
Given:
The digits 1, 2, 3, 4,3, 2, 1
The total number ofdigits are 7.
There are 4 odd digits1,1,3,3 and 4 odd places (1,3,5,7)
So, the odd digits canbe arranged in odd places in n!/ (p! × q! × r!) = 4!/(2! 2!) ways.
The remaining evendigits 2,2,4 can be arranged in 3 even places in n!/ (p! × q! × r!) = 3!/2!Ways.
Hence, the totalnumber of digits = 4!/(2! 2!) × 3!/2!
= [4×3×2×1×3×2×1] /(2! 2! 2!)
= 3×2×1×3×1
= 18
Hence, the number ofways of arranging the digits such odd digits always occupies odd places isequals to 18.