How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so

1.	How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?
Answer» Given as The digits 1, 2, 3, 4, 3, 2, 1 Total number of digits are 7. Here's 4 odd digits 1,1,3,3 and 4 odd places (1,3,5,7) Therefore, the odd digits can be arranged in odd places in n!/ (p! × q! × r!) = 4!/(2! 2!) ways. And the remaining even digits 2,2,4 can be arranged in 3 even places in n!/ (p! × q! × r!) = 3!/2! Ways. Thus, the total number of digits = 4!/(2! 2!) × 3!/2! = [4 × 3 × 2 × 1 × 3 × 2 × 1] / (2! 2! 2!) = 3 × 2 × 1 × 3 × 1 = 18 Thus, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.

How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Answer»

Given as

The digits 1, 2, 3, 4, 3, 2, 1

Total number of digits are 7.

Here's 4 odd digits 1,1,3,3 and 4 odd places (1,3,5,7)

Therefore, the odd digits can be arranged in odd places in n!/ (p! × q! × r!) = 4!/(2! 2!) ways.

And the remaining even digits 2,2,4 can be arranged in 3 even places in n!/ (p! × q! × r!) = 3!/2! Ways.

Thus, the total number of digits = 4!/(2! 2!) × 3!/2!

= [4 × 3 × 2 × 1 × 3 × 2 × 1] / (2! 2! 2!)

= 3 × 2 × 1 × 3 × 1

= 18

Thus, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.