Question

How many positive odd numbers can be formed from the digits 2, 3, 4, 5, and 6 if no digit is to be
repeated in a given number

Answer 1

HEya mate

here are the possible no.

1. 243

2. 423

3. 263

4. 623

5. 463

6. 643

7. 245

8. 425

9. 265

10. 725

hence: 10

Answer 2

Answer:

(i) Assuming all those numbers have to be used, then the answer is 48

(ii) Assuming one or more of those digits can be ommited (i.e we can form 4 digit, 3 digit, 2 digit and 1 digit numbers), then the answer is 130

Step-by-step explanation:

(i) For odd numbers the digit can only end in 3 or 5 (the only odd numbers here). Therefore there are 2 possible choices for ones place.

For each of those 2 choices, there are 4 choices for the tens place, and 3 choices for hundreds place, and so on.

∴ Total combinations = 1*2*3*4*2 (Imagine the digit, _ _ _ _ _ )

⇒ 48

(ii) The same process as above but apply it for 4 digit, 3 digit, 2 digit and 1 digit numbers. Add all of them up, you get 130.