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
Answers
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:
(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.