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Hello, I'm new to this forum, I'm not really sure if this is the

  1. Jul 29, 2011 #1
    Hello, I'm new to this forum, I'm not really sure if this is the correct board to post it on. But I've got a question, I believe regarding exponents and percentages.

    The problem I've got is that I have a number of 250 different items, and I wanted to know what the odds of randomly selecting all 250 items in order were. The way I was calculating was by multiplying 250 to the power of 250 thinking that would list all the potential outcomes. But I'm not sure if that's correct and I get an enormous number that I can't even deal with.

    Thanks in advanced!
     
  2. jcsd
  3. Jul 29, 2011 #2
    Re: Question?

    There are

    [itex] 250! = 250 \cdot 249 \cdot 248 \cdot \cdots 4 \cdot 3 \cdot 2 \cdot 1 [/itex]

    ... ways of picking up or arranging 250 different objects.

    This would be the number of outcomes. The probability of selecting all the items in a given order would therefore be

    [itex] \frac{1}{250!} [/itex]
     
    Last edited: Jul 29, 2011
  4. Jul 29, 2011 #3
    Re: Question?

    Thank's, I'm a simpleton when it comes to this type of calculations. So I just divide 1 by 250?
     
  5. Jul 29, 2011 #4
    Re: Question?

    No, that's 250! , meaning 250 factorial.
    In general n! = n(n-1)(n-2)..........1 i.e. you multiply all the numbers from 1 to n.

    As Isak BM said, you will have 250! potential outcomes
     
  6. Jul 29, 2011 #5
    Re: Question?

    Oh, I see. So how would I calculate it on a calculator... Sorry but as I sad, I'm really bad at math and trying to get this problem solved lol. Thanks for your help.
     
  7. Jul 29, 2011 #6
    Re: Question?

    You probably can't calculate it on a calculator. Many calculators don't do factorials. Even those that do will typically underflow on this calculation -- they aren't capable of representing a number that small. It's

    [tex]
    \frac{1}{323285626090910773232081455202436847099484371767378066674794242711223747555111209488817915371028199450928507353189432926730931712808990827910302790712819216765272401892647332180411862610068329253651336789308956993571353017504051317876007724793306540233900616482555224881943672586057399222641254832982204849137721776650641276858807153128978777629519139908443774787025891729732551502832417873206581884820624785826580884882554880000000000000000000000000000000000000000000000000000000000000}[/tex]

    [tex]\approx 3.093239907049846 \times 10^{-493}[/tex]
     
    Last edited: Jul 29, 2011
  8. Jul 29, 2011 #7
    Re: Question?

    So if I were to calculate 250250 is that correct? Then the odds would be 1 in that enormous number you mentioned?
     
  9. Jul 29, 2011 #8
    Re: Question?

    No, that number is 250!. 250250 is actually a far bigger number.
     
  10. Jul 29, 2011 #9
    Re: Question?

    Ok, I see. The number I was trying to calculate is actually 260, I put it in an online calculator and it gave me this number:

    3.830 195 860 836 169 235 117 497 985 604 491 875 279 556 752 309 096 960 191 300 817 480 651 475 135 399 533 485 285 838 275 429 773 913 773 383 359 294 010 103 333 339 344 249 624 060 099 745 511 339 849 626 153 802 980 398 232 848 965 472 622 820 196 848 860 832 049 579 523 313 702 327 662 760 125 732 592 551 956 622 024 712 475 139 889 122 106 940 319 324 041 688 318 583 612 166 708 334 763 727 216 738 353 107 304 842 707 002 261 430 265 483 385 206 376 839 110 078 156 900 663 427 220 806 900 528 365 808 580 136 352 143 713 956 803 295 894 115 605 151 395 493 267 411 709 188 354 023 557 693 44e+516

    So are the odds the above number to 1? Thanks again.
     
  11. Jul 29, 2011 #10
    Re: Question?

    That looks about right.
     
  12. Jul 29, 2011 #11
    Re: Question?

    Ok thanks, I know it's a stupid question, but why is it 3. decimal?
     
  13. Jul 29, 2011 #12
    Re: Question?

    http://en.wikipedia.org/wiki/Scientific_notation" [Broken].
     
    Last edited by a moderator: May 5, 2017
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