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.

Question

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.

parthkohli · Answer

Hint: the first and last are fixed values.

parthkohli · Answer

The solution would be posted in a while.

slaaibak · Answer

9 * 26^8 * 10

parthkohli · Answer

How did you get the answer? explain my friends

maheshmeghwal9 · Answer

satellite73 but don't know combination

aravindg · Answer

26^8 *10

aravindg · Answer

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

slaaibak · Answer

You are forgetting there are 9 possible arrangements.

parthkohli · Answer

lol but satellite73 is correct =))

aravindg · Answer

i waas going to xplain?

parthkohli · Answer

9 possible rearrangements?

parthkohli · Answer

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

aravindg · Answer

@ParthKohli  bt pls post the soln only after sufficient time

parthkohli · Answer

I have deleted it.

kropot72 · Answer

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

aravindg · Answer

nah ..nw evryone have read it :(

parthkohli · Answer

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

slaaibak · Answer

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

parthkohli · Answer

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

parthkohli · Answer

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

parthkohli · Answer

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

slaaibak · Answer

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.

parthkohli · Answer

Sorry for that sir.

callisto · Answer

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

parthkohli · Answer

@Callisto can you give a medal to slaibak?

aravindg · Answer

i will :)

parthkohli · Answer

Thanks :)

callisto · Answer

You can give the medal to slaibak :)

aravindg · Answer

ya @Callisto  is going to give me :P