Question

Find the total number of nine-digit numbers that can be formed using the digits \[2,2,3,3,5,5,8,8,8\] so that the odd digits occupy the even places.

Answer 1

Hint: Use the method of permutation and combination. A permutation is the number of ways in which objects from a set may be selected, generally without replacement to form a subset, whereas combination is the method of selection where the order of selection is a factor.
In this question, we need to determine the total number of nine-digit numbers that can be formed so that the odd digits occupy the even places. For this, we need to follow the concept of combinations by separating the odd digits and even digits and placing them one by one.


Complete step by step solution:
Given digits are \[2,2,3,3,5,5,8,8,8\]
Here a total of 5 even number which are \[2,2,8,8,8\] and a total of 4 odd numbers \[3,3,5,5\]
Their numbers are to be arranged in a nine-digit numbers
\[\underline O \underline E \underline O \underline E \underline O \underline E \underline O \underline E \underline O \]
In a nine-digit number, there are 5 odd digit places and 4 even digit places
So the probability of placing an odd digit (2 no’s of 3 and 2 no’s of 5) in 4 even digit place is
\[
   = \dfrac{{4!}}{{2!2!}} \\
   = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \\
   = 6 \\
 \]
Now the probability of placing an even digit (2 no’s of 2 and 3 no’s of 8) in 5 odd digit place is
\[
   = \dfrac{{5!}}{{2!3!}} \\
   = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}} \\
   = 10 \\
 \]
Hence the probability of placing the total number of nine digits numbers using the digits \[2,2,3,3,5,5,8,8,8\] is given as:
 \[
   = \dfrac{{4!}}{{2!2!}} \times \dfrac{{5!}}{{2!3!}} \\
   = 6 \times 10 \\
   = 60ways \\
 \]


Note: In this question, students must take care that whenever an odd digit is to be placed, it must be placed at the even places, vice-versa does not hold true here in this case.