To solve the problem of the number of 9-digit numbers that can be formed using all the digits of the number 123412341 such that the even digits occupy only even places, follow these steps:
Identify the digits of the number 123412341: {1, 2, 3, 4, 1, 2, 3, 4, 1}.
Even digits available are {2, 4, 2, 4}, and they can only be placed in the even-indexed positions: 2nd, 4th, 6th, and 8th positions.
The odd digits available are {1, 1, 3, 3, 1}, and these can be placed in the remaining positions: 1st, 3rd, 5th, 7th, and 9th positions.
First, arrange the even digits in even positions:
Since the even digits are 2, 4, 2, 4, we need to arrange {2, 4, 2, 4} in 4 even slots.
This arrangement includes repeating digits, hence the permutations are calculated as follows:P = \(\frac{4!}{2!2!}\) = 6.
Next, arrange the odd digits in odd positions:
The odd digits are {1, 1, 3, 3, 1}. We arrange these in 5 odd slots.
The permutations considering the repeated digits are:P = \(\frac{5!}{3!2!}\) = 10.
The total number of permutations is obtained by multiplying the permutations of both even and odd arrangements:Total = 6 × 10 = 60.
Finally, verify to ensure the computed number (60) falls within the given range, 60 to 60, confirming it is correct.
Thus, the number of 9-digit numbers formed under the given conditions is 60.
