Question

Find the number of six-digit odd numbers that are greater than 60000 and can be formed with the digits $ 5,6,7,8,9,0 $ when repetition is not allowed?
A. 150
B. 120
C. 480
D. 240

Answer 1

Hint: We first find the conditions and use them to find the number of six-digit odd numbers that can be formed with the digits $ 5,6,7,8,9,0 $ when repetition is not allowed. Then we find the condition for those numbers being greater than 60000.

Complete step by step solution:
We have to find the number of six-digit odd numbers that are greater than 60000 and can be formed with the digits $ 5,6,7,8,9,0 $ when repetition is not allowed.
Odd numbers can be created when the end digit of the number is an odd number.
The numbers are being created with $ 5,6,7,8,9,0 $ .
Odd numbers will be when the end digit is one of the numbers $ 5,7,9 $ .
We have to create six-digit odd numbers and that’s why we can’t use 0 as the first digit.
There are six places to fill up. We start by filling up the last place first. There are three options $ 5,7,9 $ for the last place.
We can take any one of them. Then we fill up the first place where we can't use the digit used in the last place and also can’t use 0. The number also has to be greater than 60000 which means we can’t use 5 as the first placed digit.
So, for the first place we have 4 options.
For the rest of the 4 places, we have 4 options without any restriction.
The middle places can be filled up in $ 4! $ ways.
Therefore, the total number of ways the numbers can be created is $ 4!\times 3\times 4 $ ways.
Total number will be $ 4!\times 3\times 4=288 $ .
Now we included the case where we have the numbers with 5 being the first digit.
We exclude those numbers. We find the numbers saying with 5.
The first place is fixed with 5. There will be 2 options for the last position.
For the middle places there will be $ 4! $ ways.
Therefore, the total number of ways the numbers can be created is $ 2\times 4!=48 $ ways.
So, the actual number is $ 288-48=240 $ .
So, the correct answer is “Option D”.

Note: We can start the problem by taking the different conditions for the last placed digits. We take 5 as the last digit and we have normal conditions and for other cases 3 options for the first placed digit.