Question

how many number 7 digits can be formed with the digits 1,2,3,4,3,2,1 in which odd number always occupy odd places

Answer 1

$4 \times 3 \times 2 \times 3 \times 2 = 24 \times 6 = 144$

Answer 2

Final Answer : 18

Steps:
1) We have 4 odd places and 3 even places.
Odd values - 1,1,3,3
There are two 1's and two 3's.

These can be arranged in 4!/(2!*2!) =6ways

2) Even numbers are 2,2,4
There are two 2's.
No. of arrangements = 3!/2! = 3ways



Since, these two both have to be satisfied, so
required no. of arrangements = 6 * 3 = 18ways