Question

How many different nine digit numbers can be formed using the digits 2, 2, 3, 3, 5, 5, 8, 8, 8 by rearranging its digits so that the odd digits occupy even positions?
1.7560
2.180
3.16
4.60

Answer 1

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.

Complete step-by-step answer:
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.
\[X - X - X - X - X\]
Here, symbol \[\left( - \right)\] is for the even places and \[\left( X \right)\] is for the odd places of the digit number.
The digits which are even are 2, 2, 8, 8 and 8.
Number of even digits \[ = 5\]
The digits which are odd are 3, 3, 5 and 5.
Number of odd digits \[ = 4\]
We have to arrange the odd digits in even places.
Number of ways to arrange the odd digits in 4 even places \[ = \dfrac{{4!}}{{2! \times 2!}}\]
On finding the value of the factorials, we get
Number of ways to arrange the odd digits in 4 even places \[ = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}\]
On further simplification, we get
Number of ways to arrange the odd digits in 4 even places \[ = 6\]
Now, we have to arrange the even digits in odd places.
Number of ways to arrange the even digits in 5 odd places \[ = \dfrac{{5!}}{{2! \times 3!}}\]
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places \[ = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}}\]
On further simplification, we get
Number of ways to arrange the even digits in 5 odd places \[ = 10\]
Total number of 9 digits number \[ = 6 \times 10 = 60\]
Hence, the required number of 9 digit numbers \[ = 60\]
Therefore, option D is the correct answer.

Note: 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.