Askiitians Tutor Team
Last Activity: 5 Months ago
To find the number of numbers that can be formed using the digits
1
,
1
,
2
,
2
,
2
,
2
,
3
,
4
,
4
1,1,2,2,2,2,3,4,4 where the odd digits (1, 3) occupy even places (2nd, 4th, 6th, etc.), you can follow these steps:
First, let's calculate the total number of permutations of all the digits. There are 9 digits in total, but some of them are repeated. You can calculate this using permutations with repetition:
Total permutations = 9! / (2! * 4!) = 9 * 8 * 7 * 6 * 5 * 4! / (2 * 4!) = 9 * 8 * 7 * 6 / 2 = 1512
Now, let's consider the even places (2nd, 4th, 6th, etc.) where odd digits (1, 3) need to be placed. There are 4 even places, and you have 2 odd digits (1, 3) to choose from.
Number of ways to choose the even places for odd digits = C(4, 2) = 6
For each of these combinations of even places, you can arrange the odd digits (1, 3) among themselves in 2! ways.
Number of ways to arrange odd digits = 2!
For the remaining places (1st, 3rd, 5th, etc.), you have the even digits (2, 2, 2, 2, 4, 4) to choose from, which can be arranged in
6
!
6! ways. However, the digit 2 appears four times and the digit 4 appears twice, so you need to account for the permutations of these repeated digits:
Number of ways to arrange even digits =
6
!
6! / (
4
!
4! *
2
!
2!)
Now, you can calculate the total number of numbers where odd digits occupy even places:
Total numbers = (Number of ways to choose even places for odd digits) * (Number of ways to arrange odd digits) * (Number of ways to arrange even digits)
Total numbers = 6 * 2! * (
6
!
6! / (
4
!
4! *
2
!
2!))
Calculate this expression to find the total number of such numbers where odd digits occupy even places.