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# How many four-digit odd numbers can be formed using the digits 0, 2, 3, 5, 6, 8 (each digit occurs only once) ?A. 64B. 72C. 86D. 96

Last updated date: 18th Jun 2024
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Hint:- Any number cannot start with zero. So, zero cannot be in first position. And for a number to be odd its last digit must be odd. So, we had to count the number of possible digits for each position. And the product of that will be the answer.

As we know that there are six digits given and that were 0, 2, 3, 5, 6 and 8.
But we had to form a four-digit number.
As the four-digit number must be odd. So, the last digit of the number must be of (i.e. 3 or 5).
So, the possible number of digits for the last position will be 2.
Now if the first digit of any number is 0, then the number of digits in a number is decreased by 1.
So, zero cannot be in first position.
So, possible digits for first position will be 2, 6, 8 and one of 5 and 3.
So, the possible number of digits for first position will be 4.
Now as we know that each digit can occur only once. So, the number of digits left will be 3 + 1(zero) = 4
So, the possible number of digits for second position will be 4.
Now there will be 3 digits left.
So, the possible number of digits for third position will be 3.
So, the total number of four-digit numbers that can be formed using digits 0, 2, 3, 5, 6 and 8 will be = $4 \times 4 \times 3 \times 2 = 96$
Hence, the correct option will be D.

Note:- Whenever we come up with up with this type of problem then first we had to place the digits on the position for which condition is given like here we had to find odd number so we placed digit at fourth position and after that start from first position and never place zero at the first position because number will become three-digit number. Each time decrease the number of left digits by one. This will be the easiest and efficient way to find the solution of the problem.