By using the permutation and combination techniques, i have attempt to solve this problem and i would like to know if where i did it wrong
how many ways to choose $12$ donuts from a store that offers thirty varieties?
here is my computation
$\Large{\binom{12+30-1}{12}= \binom{41}{12}= \frac{41!}{12!(41-12)!}}$ by using the formula $\Large\binom{n}{k}=\frac{n!}{k!(n-k)!}$
Could you tell me if i did it wrong or right??
Thank you
sincerely
Your answer is correct - assuming the following things that you did not state:
1) The order in which the donuts are chosen does not matter.
2) You are allowed to choose several donuts of the same type.
If order didn't matter and we didn't allow repetition, then the correct answer would be $\binom{30}{12}$.
If we allowed repetition and order mattered, then the correct answer would be $30^{12}$.
And if the order mattered and repetitions were not allowed, then the correct answer would be $(30)_{12}$; if you haven't seen this notation before, it is the falling factorial, and is defined by $$ (a)_b=a(a-1)(a-2)\cdots(a-(b-1))=\frac{a!}{(a-b)!}. $$
We have $30$ types of doughnut. Label the types $1$, $2$, and so on up to $30$. We represent our doughnut choice as a sequence of $0$'s and $1$'s as follows.
First write down as many $1$'s as the number of Type $1$ doughnuts we buy. (Perhaps we write down no $1$'s at this stage.) Then write down a separator $0$. Then write down the number of type $2$ doughnuts we buy, then a $0$ as separator, then the number of Type $3$ doughnuts we buy, and so on. So we will end up with $29$ $0$'s and $12$ $1$'s.
Every purchase determines a bit-string of length $29+12$ with $12$ $1$'s. Conversely, every such bit string uniquely codes a purchase.
The number of such bit strings is $\dbinom{41}{12}$. That is precisely what you got.
