## ParthKohli 3 years ago OpenStudy in real life problem: A user has 9 letters and then 2 numbers in his/her username. If you know that the first letter is 's' and the last number is '3', then how many combinations are possible for the name? Assume that all letters are small.

1. ParthKohli

Hint: the first and last are fixed values.

2. ParthKohli

The solution would be posted in a while.

3. slaaibak

9 * 26^8 * 10

4. ParthKohli

How did you get the answer? explain my friends

5. maheshmeghwal9

satellite73 but don't know combination

6. AravindG

26^8 *10

7. AravindG

@ParthKohli why did u post the answer so fast ??

8. slaaibak

You are forgetting there are 9 possible arrangements.

9. ParthKohli

lol but satellite73 is correct =))

10. AravindG

i waas going to xplain?

11. ParthKohli

9 possible rearrangements?

12. ParthKohli

Dw lol this was an easy problem.. I'll post a much more difficult one after this

13. AravindG

@ParthKohli bt pls post the soln only after sufficient time

14. ParthKohli

I have deleted it.

15. kropot72

25P8 * 10 = 25 * 24 * 23 * 22 * 21 * 20 * 19 * 18 * 10

16. AravindG

nah ..nw evryone have read it :(

17. ParthKohli

No wait... letters can repeat.. I didn't say they can't

18. slaaibak

# x x x x x x x x is not the same as x # x x x x x x x

19. ParthKohli

And then 2 numbers. I guess I framed the problem wrong.

20. ParthKohli

It's like s x x x x x x x x # 3

21. ParthKohli

I had taken satellite as an example, I'm sorry I framed it wrong though

22. slaaibak

oh... yeah you should state clearly that the numbers appear at the end, that was a bit ambiguous.. if the numbers should be in the last two spots, its 26^8 * 10, yes.

23. ParthKohli

Sorry for that sir.

24. Callisto

number of English alphbet = 26 number of number = 10 The first and the last one is fixed that is s _ _ _ _ _ _ _ 3 Where _ can be a letter or a number Case one: the number is the second one, that is s (number) (letter)x6 3 Number of combinations = 1 times 10 times 26^6 times 1 = 10 x 26^6 Case two, the number is the third one, that is s (letter) (number) (letter)x5 3 = 1 times 26 times 10 times 26^5 times1 = 10 x 26^6 ... Case seven, the number is the 8th one, that is s (letter)x6 (number) 3 = 1 times 26^6 times 10 times 1 = 26^6 times 1 Add all the cases, P required = 7 x 10 x 26^6 This must be wrong... I hardly solve a probability/combination question correctly

25. ParthKohli

@Callisto can you give a medal to slaibak?

26. AravindG

i will :)

27. ParthKohli

Thanks :)

28. Callisto

You can give the medal to slaibak :)

29. AravindG

ya @Callisto is going to give me :P