How many distinct odd 4-digit numbers can be written with the digits 1, 2, 3 and 4 if no digit may be used more than once?
4! ==24 permutations
Each permutation begins and ends with one of the 4 digits this many times:
24 / 4 ==6 permutations
Since you have two odd numbers (1 and 3), therefore you should have:
6 x 2 ==12 Odd and distinct permutations as follows:
1243 , 1423 , 2143 , 2341 , 2413 , 2431 , 3241 , 3421 , 4123 , 4213 , 4231 , 4321 , Total = 12 such odd & distinct permutations.
+37147 4 choices for first digit
3 choices for second digit
2 choices for third digit
1 choice for last digit 4 * 3 * 2* 1 = 24 combos TOTAL ===> Half of these will be odd so 12 odd numbers can be made.
+11 Hello,
To form an odd 4-digit number using 1, 2, 3, and 4, the units digit must be 1 or 3.
Case 1: The units digit is 1. We have 3 choices for the thousands digit (2, 3, or 4) and 2 choices for the hundreds digit (the remaining digits after the thousands digit and the units digit have been chosen). After the thousands and hundreds digits have been chosen, there is only 1 choice for the tens digit. Therefore, there are 3×2×1 = 6 ways to form an odd 4-digit number with 1 as the units digit.
Case 2: The units digit is 3. We have 3 choices for the thousands digit (2, 3, or 4) and 2 choices for the hundreds digit. After the thousands and hundreds digits have been chosen, there is only 1 choice for the tens digit. Therefore, there are 3×2×1 = 6 ways to form an odd 4-digit number with 3 as the units digit.
In total, there are 6+6=12 distinct odd 4-digit numbers that can be formed using the digits 1, 2, 3, and 4 if no digit may be used more than once. Nexusiceland.co.uk
