How many odd four digit numbers can be formed from the digits 1,2,3 and 4?write the possible odd numbers?

Answer 1

(A) if the digits can't be re-used:

For the number to be odd, an odd digit must be the last digit. There are only two odd digits (1 and 3). This leaves three other digits in any other order (six possibilities) for the other three digits:

Ending with 1:
2341 2431 3241 3421 4231 4321
Ending with 3:
1243 1423 2143 2413 4123 4213

There are 12 total

(B) If the digits can be re-used:

4 possibilities for the first digit
4 possibilities for the second digit
4 possibilities for the third digit
2 possibilities (1 and 3 only) for the fourth

4 * 4 * 4 * 2 = 128

1111 1121 1131 1141 1211 1221 1231 1241
(etc.)

If they want you to write out all of the possibilities, it's almost certainly (A), where the digits can't be re-used and all four must be used in the number.

Answer 2

The last digit must be odd, so it can either be 1 or 3, 2 possibilities. The next digit can be any of the other 3. The next digit can be any of the other 2. The last digit is the remaining digit.

2 x 3 x 2 x 1 = 12

1243
1423
2143
2341
2413
2431
3241
3421
4123
4213
4231
4321

Answer 3

3+7+9

Answer 4

use combinatorics. the possible unit digits are 1 and 3. the tenths place digit can be any one of the 3 remaining numbers. the hundredsth place digit can be any one of the remaining 2 numbers. the thousanths place digit will be the remaing number.
the possible out comes is 2 x 3 2 x 1= 12

Answer 5

it either ends in 1 or 3. for either, there will be 3*2*1 = 6 permutations of the preceeding numbers. Thus a total of 12 such numbers.

Answer 6

well you need all combinations which end with 1 or 3:

1243
1423
2413
2143
4123
4213

2341
2431
3241
3421
4231
4321
So the answer is 12 combinations.

Answer 7

2341
2431
3241
3421
4231
4321
2413
2143
4123
4213
1423
1243

total 12

Answer 8

1423
2341
3241
4123