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ParthKohli

  • 2 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.

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  1. ParthKohli
    • 2 years ago
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    Hint: the first and last are fixed values.

  2. ParthKohli
    • 2 years ago
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    The solution would be posted in a while.

  3. slaaibak
    • 2 years ago
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    9 * 26^8 * 10

  4. ParthKohli
    • 2 years ago
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    How did you get the answer? explain my friends

  5. maheshmeghwal9
    • 2 years ago
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    satellite73 but don't know combination

  6. AravindG
    • 2 years ago
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    26^8 *10

  7. AravindG
    • 2 years ago
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    @ParthKohli why did u post the answer so fast ??

  8. slaaibak
    • 2 years ago
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    You are forgetting there are 9 possible arrangements.

  9. ParthKohli
    • 2 years ago
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    lol but satellite73 is correct =))

  10. AravindG
    • 2 years ago
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    i waas going to xplain?

  11. ParthKohli
    • 2 years ago
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    9 possible rearrangements?

  12. ParthKohli
    • 2 years ago
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    Dw lol this was an easy problem.. I'll post a much more difficult one after this

  13. AravindG
    • 2 years ago
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    @ParthKohli bt pls post the soln only after sufficient time

  14. ParthKohli
    • 2 years ago
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    I have deleted it.

  15. kropot72
    • 2 years ago
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    25P8 * 10 = 25 * 24 * 23 * 22 * 21 * 20 * 19 * 18 * 10

  16. AravindG
    • 2 years ago
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    nah ..nw evryone have read it :(

  17. ParthKohli
    • 2 years ago
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    No wait... letters can repeat.. I didn't say they can't

  18. slaaibak
    • 2 years ago
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    # x x x x x x x x is not the same as x # x x x x x x x

  19. ParthKohli
    • 2 years ago
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    And then 2 numbers. I guess I framed the problem wrong.

  20. ParthKohli
    • 2 years ago
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    It's like s x x x x x x x x # 3

  21. ParthKohli
    • 2 years ago
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    I had taken satellite as an example, I'm sorry I framed it wrong though

  22. slaaibak
    • 2 years ago
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    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
    • 2 years ago
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    Sorry for that sir.

  24. Callisto
    • 2 years ago
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    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
    • 2 years ago
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    @Callisto can you give a medal to slaibak?

  26. AravindG
    • 2 years ago
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    i will :)

  27. ParthKohli
    • 2 years ago
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    Thanks :)

  28. Callisto
    • 2 years ago
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    You can give the medal to slaibak :)

  29. AravindG
    • 2 years ago
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    ya @Callisto is going to give me :P

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