Question

Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9 so that odd positions are occupied by odd digits.

Answer 1

To do this question, we will first see how many digits are there in the number and hence how many odd places are there. Then we will count the number of odd numbers among the given digits and find the number of ways in which they can be arranged in the odd places. Then we will find the number of ways of arranging the left over digits in the left over positions. Then we will multiply both the results and get our final answer as both of the arrangements have to happen simultaneously.

Complete step by step answer:
Here, we have been given the digits 3, 4, 5, 6, 7, 8 and 9. Now, we can see that they are a total of 7 digits. Thus, all the numbers formed will be 7 digit numbers.
Now, we know that in any 7 digit number, there are 4 odd positions, namely- first, third, fifth and the seventh position.
Now, among the given digits, we can see that the odd numbers are 3, 5, 7 and 9. We can see that they are also 4 in number and so are the odd positions.
Hence, these 4 numbers have 4 positions and the rest 3 numbers can be filled up in any of the left over positions, namely- second, fourth and sixth.
Now, we know that the number of ways of arranging ‘n’ different objects at ‘n’ different positions is given by ‘n!’.
Thus, the number of ways arranging the odd numbers at positions is given as:

& 4! \\\ & \Rightarrow 24 \\\ \end{aligned}$$ Similarly, the number of ways of arranging the even numbers at the leftover positions is given as: $\begin{aligned} & 3! \\\ & \Rightarrow 6 \\\ \end{aligned}$ Now, we know that the arrangement of the even and the odd numbers have to be done simultaneously and we also know that the number of ways two events can happen simultaneously is given by the product of the number of ways in which the individual event can happen. Hence, the number of distinct numbers that can be formed from the given digits such that they follow the given condition is given by: $\begin{aligned} & 24\times 6 \\\ & \therefore 144 \\\ \end{aligned}$ **Hence, there are a total of 144 distinct numbers which can be formed from the digits 3, 4, 5, 6, 7, 8 and 9 such that the odd digits occupy the odd places.** **Note:** Here, the number of odd digits and odd positions were equal but they can be different too. If that happens, there can be two cases- the number of odd digits is more than the number of odd places and vice-versa. In those cases, the following approach is opted (first let us assume the number of odd digits to be ‘n’ and number of odd places to be ‘r’): Case-1: n > r We will first select ‘r’ digits out of ‘n’ and arrange them and then arrange the left over with the others. Case-2: r > n We will select ‘n’ positions out of the ‘r’ positions and arrange all the odd digits on them and then arrange the remaining digits on the remaining positions.