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Probability permutations

  1. Nov 5, 2005 #1
    I just want to ask if my work is correct...
    This are the questions:
    1. In how many ways can the letters of MISSISSAUGA be arranged?
    -->6*6*6*6*6*6*6*6*6*6*6 = 362,797,056

    2. A man bought two vanilla ice creams, three chocolate cones, four strawberry cones and maple walnut cone for his 10 children. In how many ways can he distribute the flavours among the children?
    -->4*4*4*4*4*4*4*4*4*4 = 1,048,576

    3. a) how many permutations are there of the letters of the word BASKETBALL?
    --> 7*7*7*7*7*7*7*7*7*7 = 282,475,249
    b) how many of the arrangements begin with K?
    --> 1*7*7*7*7*7*7*7*7*7 = 40,353,607
    c) how may of the arrangements would the two L's be together?
    --> 1*7*7*7*7*7*7*7*7 = 5,764,801

    4. How many four digit numbers are there with the following retrictions?
    a) using the digits 1 to 8
    --> 8*7*7*7 = 2,744
    b) using the digits 0 to 7
    --> 7*6*6*6 = 1,512
    c) using only odd digits
    --> this i dont get, are you going to use 1 to 10? (i dont know)
    d) the number is odd
    --> same
    e) the number is even
    -->same
     
  2. jcsd
  3. Nov 5, 2005 #2
    4a. shouldnt it be 8P4(dont know how to write it on the comp but it looks a bit like that on the calculator). therefore 8 x 7 x 6 x 5.

    Bah my answers are wrong, I kept thinking about combinations and not permutations

    I just looked at my old maths book and I just remembered haha.
    1. is 11!/(4! x 2! x 2!)
    2. 10!/(2! x 3! x 4! x 1!)
    Just keep using the formula I wrote just down here.

    The formula I used was, (no. of letters or whatever)!/((no. of same things eg Letter S)! x (no of other same things eg Letter A)! and keep multiplying any same samples) the book writes n!/(p!q!r!...)
     
    Last edited: Nov 5, 2005
  4. Nov 5, 2005 #3
    for this, i think you are right...
     
  5. Nov 5, 2005 #4
    the formula you gave to me is very helpful... coz our teacher doesn't write anything on the board, he just keeps on talking...lol...thanks

    what if you got words with same letters with combination of different ones?
     
    Last edited: Nov 5, 2005
  6. Nov 5, 2005 #5
    I added Question 2 answer and for 4c. It should be 5P4. that is 5!/1! since there are only 5 odd numbers from 0-9 (i.e. 1,3,5,7,9). 10 is not a digit, its a combination of 2 digits.

    4d means that the last digit has to be odd.
     
  7. Nov 5, 2005 #6
    can you please give me an example, i don't quite get what you mean.
     
  8. Nov 5, 2005 #7
    thanks erzeon for helping me this time!!!
     
  9. Nov 5, 2005 #8
    i get it now, since you said that just keeping multiplying the same letters... and since it is only 1 letter, it is only 1!, so it is understood that it is one... i get waht you mean...
     
  10. Nov 5, 2005 #9
    so the answer would be 1*4*3 = 12? is it right?
     
  11. Nov 5, 2005 #10
    for 4c or 4d? , for 4c the answer is 120. it should be anyway
     
  12. Nov 5, 2005 #11
    how could that be? it is 5!/1!?
     
  13. Nov 5, 2005 #12
    (5 x 4 x 3 x 2 x 1) / 1 = 5!/1!
     
  14. Nov 5, 2005 #13
    so im right, lol
     
  15. Nov 5, 2005 #14
    how about for no. 3a, to see if i get it, ill try to answer it and then check it if is right..
     
    Last edited: Nov 5, 2005
  16. Nov 5, 2005 #15
    10!/(2!x2!x2!x1!x1!x1!x1!)
    or just 10!/((2!x2!x2!)
     
  17. Nov 5, 2005 #16
    which question is no.7?
     
  18. Nov 5, 2005 #17
    3a.....................
     
  19. Nov 5, 2005 #18
    lol...................................................
     
  20. Nov 5, 2005 #19
    yes 10!/(2! x 2! x 2!) should be correct
     
  21. Nov 5, 2005 #20
    claps, im right... its really amazing when you get to know how to solve a problem!!
     
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