Counting Principles, Permutations, and Combinations: Practice

Build the committee in two stages, choosing the women and then the men. Neither group is ordered, so each stage is a combination.

$\binom{7}{3} = 35, \qquad \binom{5}{2} = 10$

The second stage offers $10$ choices no matter which women were picked, so the multiplication principle applies.

$35 \times 10 = 350$

The value $792$ is $\binom{12}{5}$, which ignores the required mix of women and men.

If Maria must be on the committee, then one seat is already settled and the only freedom left is filling the other two seats from the remaining $9$ people. Those two seats are unordered.

$\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$

Check it against the complement. All committees number $\binom{10}{3} = 120$, and those avoiding Maria number $\binom{9}{3} = 84$, so those including her number $120 - 84 = 36$.

Sort the letters by type: one M, four I's, four S's, and two P's, and check that $1 + 4 + 4 + 2 = 11$.

All $11$ tagged letters would give $11!$ arrangements, and each actual word is produced $1!\,4!\,4!\,2!$ times, so divide.

$\frac{11!}{1!\,4!\,4!\,2!} = \frac{39{,}916{,}800}{1 \times 24 \times 24 \times 2} = \frac{39{,}916{,}800}{1{,}152} = 34{,}650$

Forgetting one of the $4!$ factors leaves $831{,}600$, which still overcounts by a factor of $24$.

Each segment joining two vertices is decided by the unordered pair of vertices it connects, so the number of such segments is

$\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28.$

Those $28$ segments include the $8$ sides of the octagon, which are not diagonals, so subtract them.

$28 - 8 = 20$

The three prizes are different, so the order of the picks matters, and no one may win twice. Fill three slots with shrinking counts.

$P(9,3) = 9 \times 8 \times 7 = 504$

The value $84$ is $\binom{9}{3}$, which would be right only if the three prizes were identical, so that only the set of winners mattered, and $729 = 9^3$ would let one contestant take all three.

Each question is a stage with $2$ options, and answering one question does not change the options available on the next.

$2^{10} = 1{,}024$

The base is the number of choices per question and the exponent is the number of questions, so $10^2 = 100$ reverses them. The value $3{,}628{,}800 = 10!$ would count orderings of the questions, which is a different problem.

The word has $5$ letters, with two L's, two E's, and one V. Tagging the letters would give $5! = 120$ arrangements, and each actual word is produced $2!\,2!\,1! = 4$ times.

$\frac{5!}{2!\,2!\,1!} = \frac{120}{4} = 30$

Dividing only by the L's, or only by the E's, would leave $60$, which still counts each word twice.

Swapping two books on the shelf gives a visibly different shelf, so the order matters, and no book can be used twice.

$P(9,4) = 9 \times 8 \times 7 \times 6 = 3{,}024$

There are exactly $4$ factors, one per shelf position, ending at $9 - 4 + 1 = 6$. The value $126$ is $\binom{9}{4}$, which counts only which books were chosen and not how they were ordered.

A game is decided by the unordered pair of teams that play it, so with $n$ teams the number of games is $\binom{n}{2}$. Set that equal to $45$.

$\frac{n(n-1)}{2} = 45 \quad\Longrightarrow\quad n(n-1) = 90 \quad\Longrightarrow\quad n^2 - n - 90 = 0$

Factor the quadratic as $(n-10)(n+9) = 0$, giving $n = 10$ or $n = -9$. A count of teams cannot be negative, so $n = 10$.

Check it: $\binom{10}{2} = 45$. With $9$ teams there would be only $\binom{9}{2} = 36$ games.

Choosing which $7$ objects to take is the same act as choosing which $3$ objects to leave behind: fix one and the other is decided. Two descriptions of one act must have the same count.

$\binom{n}{k} = \binom{n}{n-k} \quad\Longrightarrow\quad \binom{10}{7} = \binom{10}{3} = 120$

The formula agrees, since swapping $k$ with $n-k$ only swaps the two factorials in the denominator. For contrast, $\binom{10}{4} = 210$, which is a different number.

Count the opposite. Choosing any $3$ of the $9$ marbles can be done in $\binom{9}{3} = 84$ ways.

The selections that are all one color are the all-red ones and the all-blue ones, and these two cases cannot overlap, so their counts add.

$\binom{5}{3} + \binom{4}{3} = 10 + 4 = 14$

Subtract the unwanted selections from the total.

$84 - 14 = 70$

Decide each topping separately. Every one of the $8$ toppings is either on the pizza or off it, and that decision is independent of the others, so there are $2$ options at each of $8$ stages.

$2^8 = 256$

The same count arrives another way: the pizzas with exactly $k$ toppings number $\binom{8}{k}$, and summing $\binom{8}{0} + \binom{8}{1} + \cdots + \binom{8}{8}$ gives $256$, which is the fact that a row of Pascal's triangle adds up to a power of two.