Question
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?
- 16
- 36
- 60
- 180
Hint:
Here we need to find the total number of nine digit numbers that can be formed using the given digits. We will count the number of even places present for the odd digits and then we will find the number of odd places present for the even digits. Then we will find the number of ways to arrange the odd digits and then we will find the number of ways to arrange the even digits and to get the final answer, we will multiply both of them.
The correct answer is: 60
Detailed Solution
Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
The digits which are even are 2, 2, 8, 8 and 8.
The digits which are odd are 3, 3, 5 and 5.
We have to arrange the odd digits in even places.
On finding the value of the factorials, we get
On further simplification, we get
Now, we have to arrange the even digits in odd places.
On finding the value of the factorials, we get
On further simplification, we get
Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.
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Alternatively, we can use the formula for the sum of numbers as
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are respectively
