Question

Consider all numbers between 100 and 999 that have distinct digits how many of them are odd.

Answer 1

Hint: Here we will form all the possible three digit numbers from 100 to 999 such that the number at the unit place of each number so formed is odd.
Hence, we will use the concept of permutation and combination to get the desired answer.

Complete step by step answer:
We know that a number can be formed by using the following digits:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Now we have to form a three digit number between 100 and 999 such that it is odd and has distinct digits i.e. no digits in a number are the same and also the digit at unit’s place of the number is odd.
Let us make three boxes for three digits in the number.

Hundreds	Tens	Ones

Now we know that one's place can be filled only with odd numbers.
Now we know that the odd numbers from 0 to 9 are 1 , 3, 5, 7, 9
Hence, the one’s place can be filled by 5 ways i.e. the third box can be filled by 5 ways.
Now, the first box i.e. hundreds places can be filled by 8 ways.
Also, since the digits cannot be repeated
Hence, the second box i.e. the tens place can be filled by 8 ways.
Hence, the total number of ways in the three digit odd number can be formed such that it has distinct digits is:-
\[ = 8 \times 8 \times 5\]
\[ = 320\] ways

Hence there are 320 such numbers between 100 and 999 which are odd and have distinct digits.

Note:
Students should note that at hundreds place zero cannot be placed as no three digit number starts with zero and hence it can be filled by only 8 ways.
Also, students should take care that none of the digits can be repeated in a number.