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anonymous

  • 5 years ago

am confused in simple matter 4!/2! can anyone explain it plz?

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  1. anonymous
    • 5 years ago
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    4!=4*3*2*1 2!=2*1 4!/2!=4*3=12

  2. anonymous
    • 5 years ago
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    = 1*2*3*4 / 1*2 1 & 2 - cancel out so: =3*4/1=12

  3. anonymous
    • 5 years ago
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    the number of ways to pick two out of 4 is (4 2) =6

  4. anonymous
    • 5 years ago
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    right?

  5. anonymous
    • 5 years ago
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    yeah that is 4!/(2!2!)

  6. anonymous
    • 5 years ago
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    how its double 2.

  7. anonymous
    • 5 years ago
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    do you know the equation for nCr?

  8. anonymous
    • 5 years ago
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    not exactly.

  9. anonymous
    • 5 years ago
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    c means combination.

  10. anonymous
    • 5 years ago
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    when you're choosing a certain number of objects from a total you partition into a group being chosen and a group not being chosen. so n choose r we have n! total possibilities and we have to divide by (n-r)!*(r!) which is the number being chosen and the number not being chosen

  11. anonymous
    • 5 years ago
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    nCr(n,r)=n!/((n-r)!r!)

  12. anonymous
    • 5 years ago
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    do you want to try to compute 6 choose 2 and I'll tell you if it's right?

  13. anonymous
    • 5 years ago
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    4!/2! = 4x3x2!/2! = 4x3 = 12

  14. anonymous
    • 5 years ago
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    is this not same as combination..

  15. anonymous
    • 5 years ago
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    combination where order does not matter is nCr and I posted that formula above. 4!/2! is not in the standard form for a combination

  16. anonymous
    • 5 years ago
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    combination of 4 choose 2 is 4!/(2!2!)=6

  17. anonymous
    • 5 years ago
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    if we have to calculate to " how many ways can it be done" which method should be applied?

  18. anonymous
    • 5 years ago
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    it depends on the question

  19. anonymous
    • 5 years ago
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    do you have a specific problem

  20. anonymous
    • 5 years ago
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    from a grp of 7 men and 6 women, 5 persons are to be selected to form a comitte so that at least 3 men are there on comittee. in how many way can it be done.

  21. anonymous
    • 5 years ago
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    so there are three cases, you choose 3 4 or 5 men. choose three men: (7 c 3)*(6 c 2) choose 4: (7 c 4)*(6 c 1) choose all 5 men: (7 c 5) Then you would add all three cases together

  22. anonymous
    • 5 years ago
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    so basically you're choosing your men out of 7 * the remaining people you need in the comittee chosen form the total number of girls

  23. anonymous
    • 5 years ago
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    does that make sense?

  24. anonymous
    • 5 years ago
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    how do solve for (7 c 3)

  25. anonymous
    • 5 years ago
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    am the student of medicine so dont have an idea on math.

  26. anonymous
    • 5 years ago
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    I'm a graduate math student and I dont know anything about medicine haha. (7 c 3)= 7!/((7-3)!*3!)=7!/(4!3!)=35

  27. anonymous
    • 5 years ago
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    (7*6*5/3*2*1) is it right?

  28. anonymous
    • 5 years ago
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    yep

  29. anonymous
    • 5 years ago
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    but urs is different from it isnt it?

  30. anonymous
    • 5 years ago
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    no it is the same. 7!=7*6*5*4! so the 4! goes away

  31. anonymous
    • 5 years ago
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    (7*6*5/3*2*1) how it becames.....shouldnt it be like (7*6*5*4*3*2/3*2*1) ;)

  32. anonymous
    • 5 years ago
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    no. 7*6*5*4!/(4!*3*2*1) 4! cancels. 7*6*5/(3*2*1)

  33. anonymous
    • 5 years ago
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    3*2*1=6 so 6 cancels as well. 7*5=35= (7 c 3)

  34. anonymous
    • 5 years ago
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    in combination we should always use the formula n c r= n!/(n-r)*r

  35. anonymous
    • 5 years ago
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    right?

  36. anonymous
    • 5 years ago
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    if you are choosing r objects out of n and order does not matter . the two terms in the denominator are factorial as well. n!/(n-r)!r!

  37. anonymous
    • 5 years ago
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    is it necessary to write r! as well after n-r! r! comes in n-r! isnt it?

  38. anonymous
    • 5 years ago
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    anyway i got the things that i lost.thanks

  39. anonymous
    • 5 years ago
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    the equation needs both. n!/((n-r)!*r!)

  40. anonymous
    • 5 years ago
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    would you plz solve this : a man has four socks in his drawer . each socket is either black or white . the changes of him selecting a pair at random and finding that he has a white pair is 0.5. what one his chances of the pair being black?

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