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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?
fifteen cans of soup consisting of five identical cans of tomato soup, so each end consists 15/5=????? can you solve it @marceloronniel
I still have no idea at all
divide 15/3 to get 1 each end...ok
@AravindG can you explain it what am i saying to marce
15/3=?? what is the answer
can you divide 15 by 3 @marceloronniel
yes.....there is no problem about that, i only long here for the explanation but thanks anyway :)
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.
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? :)
I only learned about the simple easy problems involving permutations with repititions...I have no idea about them when it involve twists... sadly :(
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?
Yes that's where i have been familiar
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?
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?
@mayankdevnani sorry bro I was not here at that time ..seems @chihiroasleaf has everything in control now :)
the remaining 13 cans.........
yes... and what are they?
3 cans of tomato, four cans of mushroom 3 cans of celery 3 cans of vegetable
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? :)
1 201 200 ways???????
I don't count it yet..., how do you find it? :) in factorial notation?
I saw the way kropot72 did it err
ok.., it's correct... :) do you get it now?
how about letter b? even only the visualization.......
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?
so you're pointing out for mo to just multiply my answer by 4?
no.., it's adding not multiply...
so.., you have to find the number of ways at each possibility and add the result.. the sum will be the answer..
hahaha i'm i don't mean it ...was just excited to finish my task...
I got it...thank you without you...i'm lost.....:)
you're very welcome... :) good luck