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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.
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.

This Question is Closed

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Hint: the first and last are fixed values.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1The solution would be posted in a while.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1How did you get the answer? explain my friends

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0satellite73 but don't know combination

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1@ParthKohli why did u post the answer so fast ??

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.1You are forgetting there are 9 possible arrangements.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1lol but satellite73 is correct =))

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1i waas going to xplain?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.19 possible rearrangements?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Dw lol this was an easy problem.. I'll post a much more difficult one after this

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1@ParthKohli bt pls post the soln only after sufficient time

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.025P8 * 10 = 25 * 24 * 23 * 22 * 21 * 20 * 19 * 18 * 10

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1nah ..nw evryone have read it :(

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1No wait... letters can repeat.. I didn't say they can't

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.1# x x x x x x x x is not the same as x # x x x x x x x

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1And then 2 numbers. I guess I framed the problem wrong.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1It's like s x x x x x x x x # 3

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1I had taken satellite as an example, I'm sorry I framed it wrong though

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.1oh... 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1number 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
 3 years ago
Best ResponseYou've already chosen the best response.1@Callisto can you give a medal to slaibak?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1You can give the medal to slaibak :)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1ya @Callisto is going to give me :P
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