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marceloronniel

  • 2 years ago

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?

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  1. mayankdevnani
    • 2 years ago
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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

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

  3. mayankdevnani
    • 2 years ago
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    divide 15/3 to get 1 each end...ok

  4. marceloronniel
    • 2 years ago
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    huh?

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

  6. mayankdevnani
    • 2 years ago
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    15/3=?? what is the answer

  7. mayankdevnani
    • 2 years ago
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    can you divide 15 by 3 @marceloronniel

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

  9. mayankdevnani
    • 2 years ago
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    hey!!!!! lol

  10. kropot72
    • 2 years ago
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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.

  11. chihiroasleaf
    • 2 years ago
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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? :)

  12. marceloronniel
    • 2 years ago
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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 :(

  13. chihiroasleaf
    • 2 years ago
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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?

  14. marceloronniel
    • 2 years ago
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    Yes that's where i have been familiar

  15. chihiroasleaf
    • 2 years ago
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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?

  16. marceloronniel
    • 2 years ago
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    yeah

  17. chihiroasleaf
    • 2 years ago
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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?

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

  19. marceloronniel
    • 2 years ago
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    the remaining 13 cans.........

  20. chihiroasleaf
    • 2 years ago
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    yes... and what are they?

  21. marceloronniel
    • 2 years ago
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    3 cans of tomato, four cans of mushroom 3 cans of celery 3 cans of vegetable

  22. chihiroasleaf
    • 2 years ago
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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? :)

  23. marceloronniel
    • 2 years ago
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    1 201 200 ways???????

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

  25. marceloronniel
    • 2 years ago
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    I saw the way kropot72 did it err

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

  27. marceloronniel
    • 2 years ago
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    how about letter b? even only the visualization.......

  28. marceloronniel
    • 2 years ago
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    pleeease!!!! ty

  29. chihiroasleaf
    • 2 years ago
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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?

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

  31. chihiroasleaf
    • 2 years ago
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    no.., it's adding not multiply...

  32. chihiroasleaf
    • 2 years ago
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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..

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

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

  35. chihiroasleaf
    • 2 years ago
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    you're very welcome... :) good luck

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