Question

Find the number of 5-digit odd numbers that can be formed using the integers from 3 to 9 if no digit is to occur more than once in any number.
A. 1440
B. 180
C. 360
D. 720

Answer 1

Hint: In order to solve this question we have to understand the definition of odd number and then filter out the integers which can be at unit places, so that the number formed will be odd. After figuring this out, count the total number using permutations and get the answer.

Complete step-by-step answer:
In order to solve this question, we have to know what an odd number is.
Odd number is a number which is not divisible by 2, and the divisibility criteria of 2 is if the unit digit number is 0,2,4,6,8 then the number will be even and divisible by 2.
Now, the given integers from which we have to choose are 3,4,5,6,7,8,9 i.e. 7 integers.
We have to make a five digit odd number, so by using the above definition of odd number the unit digits can only be 3,5,7,9 otherwise it will be divisible by 2 and hence not odd.
So, for the unit digit we have 4 integers. Hence, unit digits can be chosen in $ ^{4}{{P}_{1}} $ i.e. 4 ways.
Now, for the remaining 4 digits, we can choose any number out of the remaining 6 as it doesn’t affect whether the number forming will be odd or even, because only the unit digit will decide which we already took care.
Hence, the remaining 4 digits can be chosen in $ ^{6}{{P}_{4}} $ i.e. $ \dfrac{6!}{(6-4)!}=\dfrac{6!}{2!}=6\times 5\times 4\times 3=360 $ ways.
So, the total number of 5-digit numbers that can be formed using the integers from 3 to 9 if no digit is to occur more than once in any number = \[360\times 4=1440\]
So, the correct answer is “Option A”.

Note: This question tests the basic understanding of permutations as the crux of the question is finding which digits we can place at the unit place and then it’s simple counting. Students found it confusing which to use $ C $ or $ P $ , but here in this question we have to include all the possible combinations (arrangement) of the chosen digits which only $ P $ can do, hence we used that.