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0
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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?

Mar 29, 2023

#1
0

3241

1243

2341

2143

3421

4123

4321

1423

Mar 29, 2023
#2
+1

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.

Mar 29, 2023
#3
+1

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.

Mar 30, 2023
#4
0

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

Mar 30, 2023
#5
0

Thank you so much everyone!

Apr 1, 2023