This is a very good question on Permutations involving rearrangement of digits to form numbers.
In most questions on permutations with digits to form numbers, it’s better to stick to the fundamental principle of counting rather than resorting to the usage of \(n_P_r\) formula. This is because, when you stick to the multiplication principle/addition principle, you’ll be able to deal with different cases more efficiently.
On the other hand, if you are dealing with permuations of people or alphabets, the \(n_P_r\) formula works better.
The questions says that the positive integers should be made of non-zero digits which are distinct. Also, the numbers should be greater than 700 and should be odd.This tells us the following things:
We can only use the digits from 1 to 9
We cannot have numbers like 111, 122 etc in our list.
We need to form numbers which look like 789, 827, 913 etc.,
The above is how this question needs to be visualized, because it will tell you what is allowed and what is not. This will then help you to tailor your approach accordingly.
Let us now break this down into three cases.
Case 1: Three digit odd numbers starting with 7. Very often, students fall into the trap of assuming that the tens digit can be filled in 8 ways because there are 8 digits available. This is mainly due to the fact that they would have got accustomed to the process of filling up the digits from left to right, in most permutation questions related to numbers. This is not correct.
Always bear in mind that you have to deal with the constraints first. Here, the constraint is that the number that we form should be an odd number. A number is an odd number when it has 1, 3, 5, 7 or 9 in the units place.
Since the number 7 has already been used to fill the hundred’s place, we cannot fill the units place with 7. This means that there are four ways of filling the units place.
Once this is done, the tens place can now be filled in 7 ways, since 2 of the 9 available digits have already been used up.
For each way of filling the unit’s place, we have 7 ways of filling the ten’s place. Therefore, for 4 ways of filling the unit’s place, there will be 28 ways of filling the unit’s and the ten’s digits and so, 28 3-digit numbers are possible which are odd and greater than 700.
Case 2 : Three digit odd numbers starting with 9. This case is exactly the same as the one with numbers starting with 7. The only exception here is that, we cannot use the digit 9 in the unit’s place, but, we can use the other 4 viz 1, 3, 5 and 7.
Therefore, we can obtain 28 3-digit numbers which are odd and greater than 900 (and hence greater than 700).
Case 3: Three digit numbers starting with 8. IN this case, we can fill up the unit’s place with any of the 5 digits i.e. 1,3,5, 7 or 9. Therefore, in this case, we will have a total of 35 numbers.
Hence, the total number of odd numbers greater than 700 = 28 + 28 + 35 = 91.
The correct answer option is B.
As you may have observed, there’s not even a single place where we used any formula to solve the question. It was based purely on logic and basic counting principles. This is why P&C questions can be a double edged sword sometimes, but there’s no doubt that they are always fun to solve.
Hope this helps!
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