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Step-by-Step Solution
Step 1: Identify the digits and the requirement
We are to form 9-digit numbers using all the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 exactly once. We want the arrangement such that all odd digits (which are 1, 1, 3) must occupy only the even positions in the 9-digit number.
Step 2: Determine the available even positions
In a 9-digit number, the digit positions can be labeled as 1 through 9. The even positions are 2, 4, 6, and 8. Hence, there are 4 even positions in total.
Step 3: Place the odd digits 1, 1, 3 in the even positions
• We must choose 3 out of these 4 even positions to place the three odd digits (1, 1, 3). The number of ways to choose these 3 positions is
$ \binom{4}{3} = 4.$
• Now, after choosing which 3 even positions to use, we arrange the digits 1, 1, and 3 among those positions. Since 1 is repeated twice, the number of ways to arrange 1, 1, and 3 is
$ \frac{3!}{2!} = 3.$
Step 4: Arrange the remaining even digits in the remaining positions
• Once the three odd digits have taken their places in 3 of the 4 even positions, we have 6 remaining digits (the four 2’s and the two 4’s) to fill the remaining 6 positions (5 odd positions + 1 leftover even position).
• The total ways to arrange these 6 digits (2, 2, 2, 2, 4, 4) is
$ \frac{6!}{4!\,2!} = 15.$
Step 5: Combine all the possibilities
• Multiply all the counts:
$ \binom{4}{3} \times \frac{3!}{2!} \times \frac{6!}{4!\,2!} = 4 \times 3 \times 15 = 180.$
Therefore, the total number of 9-digit numbers fulfilling the condition is 180.
For those three odd digit numbers 1, 1, 3 we can choose any three positions out of the four even positions.
$ \therefore $ No of ways we can choose 3 positions out of the 4 positions = 4C3
After choosing those three positions, number of ways we can arrange three odd digit numbers = 4C3 $ \times $ ${{3!} \over {2!}}$
Then the remaining 6 digits can be arrange = ${{6!} \over {2!4!}}$ ways
$ \therefore $ Total number of 9 digit numbers = 4C3 $ \times $ ${{3!} \over {2!}}$ $ \times $ ${{6!} \over {2!4!}}$ = 180
