The question is from Permutation and Combination. Another question which combines number theory with combinatorics. Our task is to find how many odd numbers can be formed under the given conditions . This section hosts a number of questions which are on par with CAT questions in difficulty on CAT Permutation and Combination, and CAT Probability.

Question 12: How many odd numbers with distinct digits can be created using the digits 1, 2, 3, 4, 5 and 6?

- 975
- 960
- 978
- 986

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=> Single digit numbers: 1, 3 or 5: Three numbers

=> Two–digit numbers: Units digit = 1, 3 or 5. For the tens’ digit, there are 5 choices {anything apart from what went into the units digit}. So, there will be 3 × 5 = 15 such numbers.

=> Three–digit numbers: Units digit = 1, 3 or 5. For the tens’ digit, there are 5 choices {anything apart from what went into the units digit}. For the 100s’ digit, there are 4 choices {anything apart from what went into the units digit or tens digit}. So, there will be 3 * 5 * 4 = 60 such numbers.

=> 4–digit numbers: There will be 3 * 5 * 4 * 3 = 180 numbers.

=> 5–digit numbers: There will be 3 * 5 * 4 * 3 * 2 = 360 numbers.

=> 6–digit numbers: There will be 3 * 5 * 4 * 3 * 2 * 1 = 360 numbers.

Total number of numbers = 360 + 360 + 180 + 60 + 15 + 3 = 978.

The question is **"How many odd numbers with distinct digits can be created using the digits 1, 2, 3, 4, 5 and 6?"**

Choice C is the correct answer.

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