Question

(permutations and combinations)

• how many numbers can be formed from the digit 1,2,3,4,3,2,1, so those odd digits are at the odd places?​

Answer 1

Step-by-step explanation:

Number of ways so that odd digits are at the odd places

= P(4, 4) × P(3, 3) / 2! 2! 3!

= 4! × 3!/ 2×2×6

=24 × 6/ 24

= 6