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marceloronniel

In a display window, a grocer wishes to put a row of fifteen cans of soup consisting of five identical cans of tomato soup, four identical cans of mushroom soup, three identical cans of celery soup, and three identical cans of vegetable soup. A.) How many displays have a can of tomato soup at each end? B.) How many displays have a can of the same kind of soup at each end?

  • one year ago
  • one year ago

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  1. mayankdevnani
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    fifteen cans of soup consisting of five identical cans of tomato soup, so each end consists 15/5=????? can you solve it @marceloronniel

    • one year ago
  2. marceloronniel
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    I still have no idea at all

    • one year ago
  3. mayankdevnani
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    divide 15/3 to get 1 each end...ok

    • one year ago
  4. marceloronniel
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    huh?

    • one year ago
  5. mayankdevnani
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    @AravindG can you explain it what am i saying to marce

    • one year ago
  6. mayankdevnani
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    15/3=?? what is the answer

    • one year ago
  7. mayankdevnani
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    can you divide 15 by 3 @marceloronniel

    • one year ago
  8. marceloronniel
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    yes.....there is no problem about that, i only long here for the explanation but thanks anyway :)

    • one year ago
  9. mayankdevnani
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    hey!!!!! lol

    • one year ago
  10. kropot72
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    Having fixed a can of tomato soup at each end of the display, there are then 3 identical cans of tomato soup, 4 identical cans of mushroom soup, 3 identical cans of celery soup and 3 identical cans of vegetable soup to rearrange in permutations. The total number of permutations of these 13 cans is given by the following: \[\frac{13!}{3!4!3!3!}\] 13! must be divided by the number of ways the available number of identical cans of each variety of soup can be arranged without making a different display.

    • one year ago
  11. chihiroasleaf
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    this problem is similar to the problem, 'how many different ways can you arrange the letters in the word 'MISSISSIPPI' ?' have you learned about this topic? :)

    • one year ago
  12. marceloronniel
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    I only learned about the simple easy problems involving permutations with repititions...I have no idea about them when it involve twists... sadly :(

    • one year ago
  13. chihiroasleaf
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    for example.., if I have words \[AAB\] first.., assume that we distinguish the letters A, I'll use subscript \[A _{1}A _{2}B\] the number of ways to arrange this letter is 3! (using permutation) Now, the number of ways to arrange the letter \[A _{1}A _{2}\] is 2! so.., if we assume that the letter 'A' is indistinguishable the number of ways to arrange the letters 'AAB' will be \[\frac{ 3! }{2! }\] can you get it so far?

    • one year ago
  14. marceloronniel
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    Yes that's where i have been familiar

    • one year ago
  15. chihiroasleaf
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    so.., in general if we have n objects with \[n _{1}\] of the first type, \[n _{1}\] of the second type, and \[n _{1}\] of the r-th type , with \[n _{1} + n_{2} + .... + n_{r} = n \] , then the number of arrangement is \[\frac{ n! }{n _{1}! n _{2}! .... n _{r} !}\] you should have been familiar with this, right?

    • one year ago
  16. marceloronniel
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    yeah

    • one year ago
  17. chihiroasleaf
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    now.., let's look at your problem... you have 15 in totals.., with 5 identical cans of tomato soup, 4 identical cans of mushroom soup, 3 identical cans of celery soup and 3 identical cans of vegetable soup Your first problem "How many displays have a can of tomato soup at each end?" you have 5 cans of tomato.., if at the end of the rows must be the tomato.., how many cans of tomato soup left? T_ _ _ _ _ _ _ _ _ _ _ _ _ T the first can and the last can in the row must be tomato (I used the letter T) so.., what cans you should arrange in the middle?

    • one year ago
  18. AravindG
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    @mayankdevnani sorry bro I was not here at that time ..seems @chihiroasleaf has everything in control now :)

    • one year ago
  19. marceloronniel
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    the remaining 13 cans.........

    • one year ago
  20. chihiroasleaf
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    yes... and what are they?

    • one year ago
  21. marceloronniel
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    3 cans of tomato, four cans of mushroom 3 cans of celery 3 cans of vegetable

    • one year ago
  22. chihiroasleaf
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    yup.., so you have 13 cans with 3 cans of tomato, 4 cans of mushroom, 3 cans of celery, and 3 cans of vegetable.., now.., how many ways can you arrange this 13 cans? :)

    • one year ago
  23. marceloronniel
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    1 201 200 ways???????

    • one year ago
  24. chihiroasleaf
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    I don't count it yet..., how do you find it? :) in factorial notation?

    • one year ago
  25. marceloronniel
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    I saw the way kropot72 did it err

    • one year ago
  26. chihiroasleaf
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    ok.., it's correct... :) do you get it now?

    • one year ago
  27. marceloronniel
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    how about letter b? even only the visualization.......

    • one year ago
  28. marceloronniel
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    pleeease!!!! ty

    • one year ago
  29. chihiroasleaf
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    the question is 'How many displays have a can of the same kind of soup at each end?' you have can of tomato(T), mushroom(M), celery (C), and vegetable (V) can you list the possibility of row with the same kind of soup at each end? for example : T _ _ _ _ _ _ _ _ _ _ _ _ _ T can you list the others?

    • one year ago
  30. marceloronniel
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    so you're pointing out for mo to just multiply my answer by 4?

    • one year ago
  31. chihiroasleaf
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    no.., it's adding not multiply...

    • one year ago
  32. chihiroasleaf
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    so.., you have to find the number of ways at each possibility and add the result.. the sum will be the answer..

    • one year ago
  33. marceloronniel
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    hahaha i'm i don't mean it ...was just excited to finish my task...

    • one year ago
  34. marceloronniel
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    I got it...thank you without you...i'm lost.....:)

    • one year ago
  35. chihiroasleaf
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    you're very welcome... :) good luck

    • one year ago
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