Question

The number of 6-digit numbers that can be made with the digits 0, 1, 2, 3, 4 and 5 so that even digits occupy odd places, is
(a). 24
(b). 36
(c).48
(d). None of these

Answer 1

Hint: Here, the even digits are 0, 2, 4 and they occupy odd places. The odd places are 1st, 3rd and 5th. But the number 0 cannot occupy the 1st place. It can be placed only in 3rd and 5th position. Similarly the odd numbers take the even position. Now, we have to calculate the number of 6- digit numbers that can be made with the digits 0, 1, 2, 3, 4 and 5 with the given condition.

Complete step-by-step solution -
Here, given the digits 0, 1, 2, 3, 4 and 5. Now, we have to make a 6-digit number with even digits occupying odd places.
In the digits 0, 1, 2, 3, 4 and 5 the even digits are 0, 2 and 4 and the odd digits are 1, 3 and 5.
We are given in the question that even digits occupy odd places. The odd places are 1st, 3rd and 5th.
Since there are 3 places, we can place the numbers 0, 2 and 4 in 3! ways. i.e.
$\begin{align}
  & 3!=1\times 2\times 3 \\
 & 3!=6 \\
\end{align}$
Hence, we can say that 0, 2 and 4 can be placed in 6 ways.
But if we place the number 0 at the 1st place, then it will have no value. That is, the number will be a 5-digit number. So, we can’t place the number 0 at the first place.
Hence, the number of ways to place 0, 2 and 4, so that the number will be a 6-digit number = 3! – 2!
3! – 2! = 6 – 2
3! – 2! = 4
 Hence there are 4 ways to place 0, 2 and 4 in the odd places for the number to be a 6-digit number.
The odd numbers 1, 3 and 5 are placed in even places, 2nd, 4th and 6th.
 The number of ways 1, 3 and 5 are placed in odd places = 3! = 6
Hence, we can say that there are 6 ways to place the numbers 1, 3 and 5 in odd places for the number to be a 6-digit number.
Now, we have to calculate the total number of ways to arrange the numbers 0, 1, 2, 3, 4, 5 and 6, so that the number is a 6-digit number.
Total number of ways = $6\times 4=24$
 Hence, we will get there are 24 6-digit numbers that can be made with the digits 0, 1, 2, 3, 4, 5 and 6, with even digits at the odd place.
Therefore, the correct answer for this question is option (a).

 Note: Here, there are 3 even numbers and 3 odd numbers in 0, 1, 2, 3, 4, 5 and 6. If 1, 2 and 4 are kept at the odd places, then 1, 3 and 5 will be kept at the even places, and you have to calculate the number of ways to place them only in even places.