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Question
How many 3-digit odd numbers can be formed from the digits 1, 2, 3, 4, 5, 6 when (i) repetition of digits is not allowed
(ii) repetition of digits is allowed?
How many 3-digit odd numbers can be formed from the digits 1, 2, 3, 4, 5, 6 when (i) repetition of digits is not allowed
(ii) repetition of digits is allowed?
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Solution
The correct option is A (i) 60, (ii) 108
(i) When repetition of digits is not allowed: Since we have to form a 3-digit odd number, thus the digit at unit's place must be odd. Hence, the unit's place can be filled up by 1, 3 or 5, that is, in 3 ways.
Now, the ten's digit can be filled up by any of the remaining 5 digits in 5 ways and then the hundred's place can be filled up by the remaining 4 digits in 4 ways.
Hence, the number of 3-digit odd numbers that can be formed =3×4×5=60.
(ii) When repetition of digits is allowed: Again, the unit's place can be filled up by 1, 3, 5. that is, in 3 ways. But the ten's and hundred's place can be filled up by any of the 6 given digits in 6 ways each. (since repetition is allowed)
Hence, the number of 3-digit odd numbers that can be formed =3×6×6=108.
(i) When repetition of digits is not allowed: Since we have to form a 3-digit odd number, thus the digit at unit's place must be odd. Hence, the unit's place can be filled up by 1, 3 or 5, that is, in 3 ways.
Now, the ten's digit can be filled up by any of the remaining 5 digits in 5 ways and then the hundred's place can be filled up by the remaining 4 digits in 4 ways.
Hence, the number of 3-digit odd numbers that can be formed =3×4×5=60.
(ii) When repetition of digits is allowed: Again, the unit's place can be filled up by 1, 3, 5. that is, in 3 ways. But the ten's and hundred's place can be filled up by any of the 6 given digits in 6 ways each. (since repetition is allowed)
Hence, the number of 3-digit odd numbers that can be formed =3×6×6=108.
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