I'm working on some combination practice problems, but I'm having a hard time wrapping my head around how to solve some of them. I'd like some feedback on my solutions and how to tackle Q3.
Question: Baker only sells sesame, poppy, sourdough, wholemeal and multigrain bagels. We would like to purchase a dozen (12).
Q1. How many combinations of bagels can be chosen if the baker has at least 12 of each type of bagel?
A: This seems to me to be a bit of a trick question with the tailing "at least 12 of each type" put there to throw the student off. The solution is the formula for combinations with repetition:
$ 12 + 5 – 1 \choose 5 – 1$ or $16 \choose 4$ which is just $1820$.
Q2. How many combinations of a dozen bagels can the baker supply if the baker has at least 12 of each type of bagel, but we want to make sure that we buy at least 2 sesame and 3 poppy bagels?
A: For this I added the restriction of 3 sesame and 2 poppy and subtracted it from the set of 12 to give 7, then used the formula for combinations -> $ 7+ 5 -1 \choose 5-1 $ or $11 \choose 4$ which equals $330$.
Q3. How many different bagel combinations of a dozen can the baker make if he only has 7 sourdough and 5 multigrain bagels left? (He has at least 12 of the poppy, sesame and white bagels)
A: This question has me stumped. I know that applying the inclusion-exclusion can solve this problem, but I'm not exactly sure how to implement the solution or which formula to use.
Any help with Q3 would be greatly appreciated.
Best Answer
You solved the first two parts of the problem correctly. The condition that there are at least $12$ bagels of each type allows you to use the formula for combinations with repetition since there at least as many bagels of each type as there are bagels in the order.
As you correctly found, there are $$\binom{12 + 5 - 1}{5 - 1} = \binom{16}{4}$$ orders when at least $12$ bagels of each type are available.
What we must do is exclude the possibilities that the baker sells more than seven sourdough bagels or more than five multigrain bagels. Notice that the baker cannot violate both of these restrictions simultaneously since $8 + 6 = 14 > 12$.
Orders with more than $7$ sourdough bagels: Such orders must contain at least $8$ sourdough bagels. Place eight sourdough bagels in the order. To complete the order, we must select four bagels from the five types of bagels. The number of ways this can be done is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 4$$ in the nonnegative integers.
Orders with more than $5$ multigrain bagels: Such orders must contain at least $6$ multigrain bagels. Place six multigrain bagels in the order. To complete the order, we must select six bagels from five types of bagels. The number of ways this can be done is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 6$$ in the nonnegative integers.
Subtracting the number of ways we could violate the restrictions from the total gives the number of possible combinations of a dozen bagels the baker could assemble.