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There are $3$ types of sandwiches, namely chicken (C), fish (F) and ham (H), available in a restaurant. A boy wishes to place an order of $6$ sandwiches. Assuming that there is no limit in the supply of sandwiches of each type, how many such orders can the boy place?

My Attempt:

I know that this is a stars and bars problem and the solution is $8 \choose 2$. But why can't it simply be $3^6$ where each of the $6$ sandwich places has $3$ choices to fill as repetition is allowed. I know I am missing something trivial.

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    $\begingroup$ Because order is not relevant. $5$ chicken followed by one ham is the same as one ham followed by $5$ chicken. $\endgroup$
    – lulu
    Apr 27, 2018 at 13:15
  • $\begingroup$ So we use this just because order is not relevant, right?? $\endgroup$ Apr 27, 2018 at 13:18
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    $\begingroup$ yes. $3^6$ would be the correct answer if we were keeping track of the order of the sandwiches. $\endgroup$
    – lulu
    Apr 27, 2018 at 13:22
  • $\begingroup$ Okay thank-you. $\endgroup$ Apr 27, 2018 at 13:31
  • $\begingroup$ In this case, we need to ask whether order matters. Does ordering chicken, fish, then ham in that order produce a different outcome than ordering fish, chicken, then ham in that order. Since it does not, we are looking at a combinations with repetition problem. $\endgroup$ Apr 27, 2018 at 14:52

1 Answer 1

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Let us take Chicken and Ham sandwich and total order of 3. You are wondering why it should not be $2^3$ and why it is ${(3+2-1)\choose (2-1)}$

The $2^3=8$ choices that you mention are

  • AAA - 3 Chicken sandwich

  • AAB - 2 Chicken and 1 Ham

  • ABA

  • BAA

  • ABB - 2 Ham and 1 Chicken

  • BAB

  • BBA

  • BBB - 3 Ham

The total is $2^3$ but the chocies are merely ${(3+2-1)\choose (2-1)}$ = ${4\choose1}=4$

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  • $\begingroup$ You are welcome!! $\endgroup$ Apr 27, 2018 at 13:33

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