A Chinese restaurant charges a standard price for its "family dinner" consisting of any four different dishes on the menu. The restaurant's advertisement claims that "over 300 different family dinners are possible". If this is true, what must be the least number of dishes listed on the menu?

Question

anonymous · Answer

any options available?

unklerhaukus · Answer

permutations or combinations?

lgbasallote · Answer

no idea

lgbasallote · Answer

but it is in that domain

lgbasallote · Answer

based on the question i would say combination

anonymous · Answer

you have the answer? so i can check

unklerhaukus · Answer

does a dinner have to include three items, ? can any of there be blank

lgbasallote · Answer

with all due respect @Omniscience if i give an answer i might get a random solution that gives the answer only by coincidence. So, i'd like to know the right solution so it might be best to keep the answer

unklerhaukus · Answer

i have mis-read the question

unklerhaukus · Answer

eleven ?

lgbasallote · Answer

how did you get 11?

lgbasallote · Answer

i know i have to do $$_n C _4 \ge 300$$
but how do i solve for n?

unklerhaukus · Answer

well taking combinations 
$$C^k_4>300$$

the smallest value of k that make the combination over 300 is k=11
$$C^{10}_4=210$$
$$C^{11}_4=330$$

unklerhaukus · Answer

um i just tried some numbers , not exactly a great method

unklerhaukus · Answer

i dont think there is a inverse for combinations,

lgbasallote · Answer

hmm makes sense...

lgbasallote · Answer

so there's no other way to solve for n?

unklerhaukus · Answer

$$^nC_4\geq300$$
$$\frac{n!}{4!(n-4)!}\geq300$$
$$\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}\geq300\times4!$$
$${n(n-1)(n-2)(n-3)}\geq300\times4!$$