Question

Example: 6
not
How many odd numbers of 4 digits can be formed out of the
digits 1, 2.......... 9 if repetition of digits is (i) not allowed (ü) allowed ?

Answer 1

Answer:

If we fix 1 at the one’s place then we will get the number of permutations as n!/(n-r)! = 6

Similarly, keeping 3 fixed at the one’s place we will get the total number of permutations as 6

Add both of them and we will get the total number of permutations (which is 6+6=12) in which 1 and 3 are at the one’s place (that is, they are odd)

Answer 2

Answer:

allowed

Step-by-step explanation:

For a number to be odd, the units digit should be either of 1,3,5,7 or 9

∴ Total number of ways of choosing units digit =5

Since, it is given that the digit may be repeated. So, we have 9 choices for ten's digit

Similarly, for rest two digits we can choose any of the 9 digits

∴ No. of ways of choosing first and second digit is also 9 for each.

∴ Total number of ways to form odd numbers is 5×9×9×9=5×9^3

Hence, the number of 4 digit odd numbers can be formed is 5×9^3

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