Neither. \({14 \choose 4}\) paths.
\({10\choose 4}\) bow ties.
\(P(10,4)\text{,}\) since order is important.
Neither. Assuming you will wear each of the 4 ties on just 4 of the 7 days, without repeats: \({10\choose 4}P(7,4)\text{.}\)
\(P(10,4)\text{.}\)
\({10\choose 4}\text{.}\)
Neither. Since you could repeat letters: \(10^4\text{.}\) If no repeats are allowed, it would be \(P(10,4)\text{.}\)
Neither. Actually, “k” is the 11th letter of the alphabet, so the answer is 0. If “k” was among the first 10 letters, there would only be 1 way - write it down.
Neither. Either \({9\choose 3}\) (if every kid gets an apple) or \({13 \choose 3}\) (if appleless kids are allowed).
Neither. Note that this could not be \({10 \choose 4}\) since the 10 things and 4 things are from different groups. \(4^{10}\text{,}\) assuming each kid eats one type of cerial.
\({10 \choose 4}\) - don't be fooled by the “arrange” in there - you are picking 4 out of 10 spots to put the 1's.
\({10 \choose 4}\) (assuming order is irrelevant).
Neither. \(16^{10}\) (each kid chooses yes or no to 4 varieties).
Neither. 0.
Neither. \(4^{10} - [{4\choose 1}3^{10} - {4\choose 2}2^{10} + {4 \choose 3}1^{10}]\text{.}\)
Neither. \(10\cdot 4\text{.}\)
Neither. \(4^{10}\text{.}\)
\({10 \choose 4}\) (which is the same as \({10 \choose 6}\)).
Neither. If all the kids were identical, and you wanted no empty teams, it would be \({10 \choose 4}\text{.}\) Instead, this will be the same as the number of surjective functions from a set of size 11 to a set of size 5.
\({10 \choose 4}\text{.}\)
\({10 \choose 4}\text{.}\)
Neither. \(4!\text{.}\)
Neither. It's \({10 \choose 4}\) if you won't repeat any choices. If repetition is allowed, then this becomes \(x_1 + x_2 + \cdots +x_{10} = 4\text{,}\) which has \({13 \choose 9}\) solutions in non-negative integers.
Neither. Since repetition of cookie type is allowed, the answer is \(10^4\text{.}\) Without repetition, you would have \(P(10,4)\text{.}\)
\({10 \choose 4}\) since that is equal to \({9 \choose 4} + {9 \choose 3}\text{.}\)
Neither. It will be a complicated (possibly PIE) counting problem.