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chen.aavaz

Member

Registered: 2016-05-05

Posts: 37

3 sandwiches with 4 ingredients.

We want to make 3 sandwiches using some choices from 4 different ingredients, but with the following rules:
None of the 4 can be used in all 3 sandwiches and, for any two sandwiches, there is at least one ingredient at both of them but not at the 3rd one. In how many ways can we make these 3 sandwiches while satisfying the constraints?

Last edited by chen.aavaz (2017-05-23 21:15:29)

Agnishom

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Registered: 2011-01-29

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Re: 3 sandwiches with 4 ingredients.

How many different sandwiches can we make?

I think you want to ask "In how many ways can we make these system of 3 sandwiches while satisfying the constraints?"

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chen.aavaz

Member

Registered: 2016-05-05

Posts: 37

Re: 3 sandwiches with 4 ingredients.

Yes, thank you!
Sorry, English is not my mother tongue!

How many different sandwiches can we make?

I think you want to ask "In how many ways can we make these system of 3 sandwiches while satisfying the constraints?"

mr.wong

Member

Registered: 2015-12-01

Posts: 252

Re: 3 sandwiches with 4 ingredients.

Hi   chen. aavaz ,

The  no.  of  ingredients  contained  in  each  sandwich  should  be 
either  2  or  3  , but  excluding  the  case  of  any  3  ingredients  in  all 
the  3  sandwiches .
Thus  the  possible  cases  may  be  divided  into  3  types :
(i)  ( 3 , 3 , 2 )  i.e.  2  sandwiches  both  with  3  ingredients  while 
       the  remaining  sandwich  with  only  2  ingredients .
(ii)  ( 3 , 2 , 2 )  , 1  sandwich  with  3  ingredients  while  the  2 
       remaining  sandwiches  both  with  2  ingredients .
(iii)  ( 2 , 2 , 2 ) , all  the  3  sandwiches  each  with  2  ingredients .

chen.aavaz

Member

Registered: 2016-05-05

Posts: 37

Re: 3 sandwiches with 4 ingredients.

@mr.wong: Thank you. So what is the total of possible ways?

Hi   chen. aavaz ,

The  no.  of  ingredients  contained  in  each  sandwich  should  be 
either  2  or  3  , but  excluding  the  case  of  any  3  ingredients  in  all 
the  3  sandwiches .
Thus  the  possible  cases  may  be  divided  into  3  types :
(i)  ( 3 , 3 , 2 )  i.e.  2  sandwiches  both  with  3  ingredients  while 
       the  remaining  sandwich  with  only  2  ingredients .
(ii)  ( 3 , 2 , 2 )  , 1  sandwich  with  3  ingredients  while  the  2 
       remaining  sandwiches  both  with  2  ingredients .
(iii)  ( 2 , 2 , 2 ) , all  the  3  sandwiches  each  with  2  ingredients .

mr.wong

Member

Registered: 2015-12-01

Posts: 252

Re: 3 sandwiches with 4 ingredients.

Hi  chen. aavaz ,

For  (i) , there  are  4 C 2 =  6  ways .

For (ii) , there  are  4 C 3  *  3 C 1  * 2 C 1  = 4 * 3 * 2 = 24  ways .

For (iii) , there  are  4 C 3 = 4  ways .

Therefore  the  total  no. of  possible  ways = 6 + 24 + 4 = 34  .

chen.aavaz

Member

Registered: 2016-05-05

Posts: 37

Re: 3 sandwiches with 4 ingredients.

@mr.wong: Thanks for the reply.
Can you please explain why (ii) is 4 C 3  *  3 C 1  * 2 C 1?
To me they are 48.
Also, if we rotate the 3 sandwiches (let's say that each sandwich is A, B and C), we also have 3!=6 different ways, so the total to me is 58*6=348

mr.wong

Member

Registered: 2015-12-01

Posts: 252

Re: 3 sandwiches with 4 ingredients.

Hi  chen. aavaz ,

Let  the  4  ingredients  be  denoted  by  p , q , r  and  s  .

For  (ii) , there  are  4 C 3 = 4  choice  for  the  sandwich  containing  3  ing. , say  p , q  and  r  . Then  one  of  the  remaining  sandwich  will  contain  1 ing. from  either  p  or  q  or  r  , say  p  ( there  are  3 C 1 = 3  ways .) together  with   s  which  is  left  behind . The  last  sandwich  must  contain  s  , together 
  with  either  q  or  r  but  not  p . ( there  are  2 C 1 = 2 ways .)
Thus  the  total  possible  ways  of  (ii) = 4 * 3 * 2 = 24 .

  I  have  assumed   that  the  order  of  the  sandwiches  will  not  be  considered ,  otherwise 
the  total  possible  ways  will  be  34 * 6 = 204 .

chen.aavaz

Member

Registered: 2016-05-05

Posts: 37

Re: 3 sandwiches with 4 ingredients.

I am getting 48 instead of 24. I agree for all the rest.

p,q,r,    p,q    r,s
p,q,r,    q,r    p,s
p,q,r,    p,r    q,s
   and then:
p,q,r,    p,s    q,s
p,q,r,    p,s    q,r
p,q,r,    p,s    r,s
then the same for
p,q,r,    q,s…
and then the same for
p,q,r,    r,s
So a total of 12 for p,q,r and therefore 48 for all 4 permutations for the first part (3C4).

mr.wong

Member

Registered: 2015-12-01

Posts: 252

Re: 3 sandwiches with 4 ingredients.

Hi  chen.aavaz ,

According  to  the  constraints  "  for any two sandwiches, there is at least one ingredient at both of them but not at the 3rd one. "  the  ways  in  your  1st , 2nd , 3rd , 5th  rows  etc. will  not  be  permitted .
E.g. in  the  1st  row , ( p,q)  and  (r,s)  have  no  common ingredient .

chen.aavaz

Member

Registered: 2016-05-05

Posts: 37

Re: 3 sandwiches with 4 ingredients.

ooops yes you are right!

Thank you!