Question

The number of odd numbers having 4 digits that can be formed from {1,2,3…,9} are:
A. 1680
B. ${}^8{P_3}$
C. ${}^8{P_4}$
D. ${}^9{P_4}$

Answer 1

Hint: Count the number of odd digits in the given digits, and fix the last digit one-by-one using these odd digits in order to have an odd number. Then count the number of choices a place in the 4-digit number has, keeping in mind that the repetition of digits is not allowed. Once you have calculated the number of ways by fixing each odd digit at the last place, one-by-one, add the result obtained in all the cases to find the total required numbers.

Complete step-by-step solution
Let us begin by determining the number of odd digits out of 1,2,3,4,5,6,7,8 and 9. It is known that the odd numbers or digits are those which are not completely divisible by 2.
Thus, 1,3,5,7,9 are the odd digits among the given digits.

Now, in order to find an odd number of 4 digits, it is necessary that the last digit of that number should be either 1 or 3 or 5 or 7 or 9. If this is not happening, then the number would become an even number, which would be wrong.

Thus, we have the following cases:

1. The last digit of the number is 1.
2. The last digit of the number is 3.
3. The last digit of the number is 5.
4. The last digit of the number is 7.
5. The last digit of the number is 9.

We will consider each case one -by-one.

Case 1:- The last digit of the number is 1.
If you fix the last digit to be 1, the choices left for the first place are 8. After placing two numbers, one on the first place and second on the last, the number of choices left for the second place is 7 and similarly the number of choices left for the third place is 6.
$\Rightarrow {N_1} = 8 \times 7 \times 6 \times 1 \\\ = 336 \\\$

Case 2:- The last digit of the number is 3.
If you fix the last digit to be 3, the choices left for the first place are 8. After placing two numbers, one on the first place and second on the last, the number of choices left for the second place is 7 and similarly the number of choices left for the third place is 6.
$\Rightarrow {N_2} = 8 \times 7 \times 6 \times 1 \\\ = 336 \\\$

Case 3:- The last digit of the number is 5.
If you fix the last digit to be 5, the choices left for the first place are 8. After placing two numbers, one on the first place and second on the last, the number of choices left for the second place is 7 and similarly the number of choices left for the third place is 6.
$\Rightarrow {N_3} = 8 \times 7 \times 6 \times 1 \\\ = 336 \\\$

Case 4:- The last digit of the number is 7.
If you fix the last digit to be 7, the choices left for the first place are 8. After placing two numbers, one on the first place and second on the last, the number of choices left for the second place is 7 and similarly the number of choices left for the third place is 6.
$\Rightarrow {N_4} = 8 \times 7 \times 6 \times 1 \\\ = 336 \\\$

Case 5:- The last digit of the number is 9.
If you fix the last digit to be 9, the choices left for the first place are 8. After placing two numbers, one on the first place and second on the last, the number of choices left for the second place is 7 and similarly the number of choices left for the third place is 6.
$\Rightarrow {N_5} = 8 \times 7 \times 6 \times 1 \\\ = 336 \\\$

Thus, find the total number of ways by adding all the obtained values in all the 5 cases.
$\Rightarrow N = {N_1} + {N_2} + {N_3} + {N_4} + {N_5} \\\ = 336 + 336 + 336 + 336 + 336 \\\ = 1680 \\\$

Thus, the number of odd numbers having 4 digits that can be formed from {1,2,3…,9} are 1680.

Hence, option (A) is the correct option.

Note: It is important to note that in this question we have assumed that repetition of numbers is not allowed, as nothing was given in the question. If repetition is allowed then it will be mentioned in the question. As, the answer would differ if we assume the repetition is allowed or not it is not allowed. Also, keep in mind that in the end you need to add the obtained ways and not to multiply them.