Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose a donuts shop has $20$ varieties of donuts. how many ways are there to choose at least two kinds of donuts in dozen donuts ?

please correct me:

this is combination with repetition question and we have total of $C\left(12+20-1,12\right)$ cases and we want to subtract it from the cases if we have one kind of donuts so this is $20$ so answer will be $C\left(31,12\right)-20$.

share|improve this question
    
Do you know why C(12+20-1,12) gives the number of combinations allowing repetitions? If you know why this formula works then you are not far from figuring out the number of ways of picking a single variety. –  Montez Dec 19 '12 at 1:45
1  
Yes, it’s fine. (But you’re not subtracting $\binom{31}{12}$ from $20$; you’re subtracting $20$ from $\binom{31}{12}$. Subtract $a$ from $b$ means $b-a$.) –  Brian M. Scott Dec 19 '12 at 1:58

1 Answer 1

up vote 2 down vote accepted

That sounds correct. It's certainly easier to subtract the outcomes you don't want (selecting only one type) than to add up the ones you do want (selecting 2, 3, ..., 12 types).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.