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adnanchowdhury

How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetition are not allowed?

  • 2 years ago
  • 2 years ago

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  1. bahrom7893
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    1 digit - 4 ways (any one can be chosen)

    • 2 years ago
  2. bahrom7893
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    2 - digits - 4*3 ways (you picked the first one, so now you have 3 to choose from)

    • 2 years ago
  3. bahrom7893
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    Waaait, isn't it 4C1

    • 2 years ago
  4. adnanchowdhury
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    I don't understand. Can you list the permutations? I can't seem to understand what the question is saying.

    • 2 years ago
  5. adnanchowdhury
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    Answer is supposed to be 64.

    • 2 years ago
  6. bahrom7893
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    For which one?

    • 2 years ago
  7. bahrom7893
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    wait im wrong.. hold up. You have to use that formula, like the number of ways of choosing some object from a set of objects with no repetitions

    • 2 years ago
  8. bahrom7893
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    ehhh... im lost.. crap i suck at probability

    • 2 years ago
  9. bahrom7893
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    Let me call amistre

    • 2 years ago
  10. adnanchowdhury
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    Can you explain what the question is asking? What does it mean when it says: 'take one, two, three and four digits from the digits 1, 2, 7 and 8'?

    • 2 years ago
  11. bahrom7893
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    im thinking: Pick one digit from 1,2,7 and 8. Like I can pick 1, or 2, or 7, or 8, so the answer's 4 ways

    • 2 years ago
  12. bahrom7893
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    yay amistre's here

    • 2 years ago
  13. adnanchowdhury
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    So would one possibility be 1278?

    • 2 years ago
  14. adnanchowdhury
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    One digit possibilities: 1, 2, 7 and 8. So there are 4 permutations for one digits...?

    • 2 years ago
  15. adnanchowdhury
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    I understand now!

    • 2 years ago
  16. adnanchowdhury
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    4 possibilities for 1 digits. 4x3 = 12 possibilities for 2 digits. And so on..

    • 2 years ago
  17. bahrom7893
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    yes that's what i was trying to say adnan

    • 2 years ago
  18. adnanchowdhury
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    Thanks! My problem was in understanding what the question was asking, but after you explained it, i understood.

    • 2 years ago
  19. amistre64
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    1, 2, 7 and 8 1 digit numbers; there are only 4 numbers so: 4 2 digit number: 12 17 18; 4 times = 12; 4P1 * 3P1 = 12 3digit number: 127 128 172 178 182 187; 6*4 = 24; 4*3*2 = 24 4 digit numbers: 1278 1287 1728 1782 1827 1872: 6*4 = 24; 4*3*2*1 = 24 4+12+24+24 = 4+12(3) = 40 i beleive

    • 2 years ago
  20. amistre64
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    lol, almost had that right

    • 2 years ago
  21. amistre64
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    60+4 if i learn to count :)

    • 2 years ago
  22. adnanchowdhury
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    yes lol

    • 2 years ago
  23. adnanchowdhury
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    Any formula i could have used to save me time?

    • 2 years ago
  24. amistre64
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    not really; since it is split up into 4 groups to count individually you still have a bit of work

    • 2 years ago
  25. amistre64
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    knowing the cPr stuff helps tho

    • 2 years ago
  26. amistre64
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    nPr ... cant type

    • 2 years ago
  27. adnanchowdhury
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    I understood the formula before. But now, I just find it confusing.

    • 2 years ago
  28. amistre64
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    yeah, its just a "structured" way to count and if you dont use it you tend to go back to the basic method of 1,2,3,4,5s

    • 2 years ago
  29. adnanchowdhury
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    I see. I work it out intuitively now - probably takes me a bit longer, but i get there eventually.

    • 2 years ago
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